mirror of
https://github.com/halfrost/LeetCode-Go.git
synced 2025-07-06 09:23:19 +08:00
Add solution 0119、1439
This commit is contained in:
YDZ
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package leetcode
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func getRow(rowIndex int) []int {
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row := make([]int, rowIndex+1)
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row[0] = 1
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for i := 1; i <= rowIndex; i++ {
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row[i] = row[i-1] * (rowIndex - i + 1) / i
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}
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return row
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}
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package leetcode
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import (
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"fmt"
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"testing"
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)
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type question119 struct {
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para119
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ans119
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}
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// para 是参数
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// one 代表第一个参数
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type para119 struct {
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rowIndex int
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}
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// ans 是答案
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// one 代表第一个答案
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type ans119 struct {
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one []int
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}
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func Test_Problem119(t *testing.T) {
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qs := []question119{
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{
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para119{3},
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ans119{[]int{1, 3, 3, 1}},
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},
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{
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para119{0},
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ans119{[]int{1}},
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},
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}
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fmt.Printf("------------------------Leetcode Problem 119------------------------\n")
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for _, q := range qs {
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_, p := q.ans119, q.para119
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fmt.Printf("【input】:%v 【output】:%v\n", p, getRow(p.rowIndex))
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}
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fmt.Printf("\n\n\n")
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}
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72
leetcode/0119.Pascals-Triangle-II/README.md
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72
leetcode/0119.Pascals-Triangle-II/README.md
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# [119. Pascal's Triangle II](https://leetcode.com/problems/pascals-triangle-ii/)
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## 题目
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Given an integer `rowIndex`, return the `rowIndexth` row of the Pascal's triangle.
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Notice that the row index starts from **0**.
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In Pascal's triangle, each number is the sum of the two numbers directly above it.
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**Follow up:**
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Could you optimize your algorithm to use only *O*(*k*) extra space?
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**Example 1:**
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```
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Input: rowIndex = 3
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Output: [1,3,3,1]
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```
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**Example 2:**
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```
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Input: rowIndex = 0
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Output: [1]
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```
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**Example 3:**
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```
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Input: rowIndex = 1
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Output: [1,1]
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```
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**Constraints:**
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- `0 <= rowIndex <= 33`
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## 题目大意
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给定一个非负索引 k，其中 k ≤ 33，返回杨辉三角的第 k 行。
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## 解题思路
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- 题目中的三角是杨辉三角，每个数字是 `(a+b)^n` 二项式展开的系数。题目要求我们只能使用 O(k) 的空间。那么需要找到两两项直接的递推关系。由组合知识得知：
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$$\begin{aligned}C_{n}^{m} &= \frac{n!}{m!(n-m)!} \\C_{n}^{m-1} &= \frac{n!}{(m-1)!(n-m+1)!}\end{aligned}$$
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于是得到递推公式：
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$$C_{n}^{m} = C_{n}^{m-1} \times \frac{n-m+1}{m}$$
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利用这个递推公式即可以把空间复杂度优化到 O(k)
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## 代码
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```go
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package leetcode
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func getRow(rowIndex int) []int {
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row := make([]int, rowIndex+1)
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row[0] = 1
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for i := 1; i <= rowIndex; i++ {
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row[i] = row[i-1] * (rowIndex - i + 1) / i
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}
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return row
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}
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```
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@ -0,0 +1,82 @@
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package leetcode
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import "container/heap"
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func kthSmallest(mat [][]int, k int) int {
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if len(mat) == 0 || len(mat[0]) == 0 || k == 0 {
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return 0
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}
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prev := mat[0]
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for i := 1; i < len(mat); i++ {
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prev = kSmallestPairs(prev, mat[i], k)
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}
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if k < len(prev) {
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return -1
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}
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return prev[k-1]
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}
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func kSmallestPairs(nums1 []int, nums2 []int, k int) []int {
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res := []int{}
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if len(nums2) == 0 {
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return res
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}
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pq := newPriorityQueue()
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for i := 0; i < len(nums1) && i < k; i++ {
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heap.Push(pq, &pddata{
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n1: nums1[i],
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n2: nums2[0],
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n2Idx: 0,
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})
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}
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for pq.Len() > 0 {
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i := heap.Pop(pq)
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data := i.(*pddata)
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res = append(res, data.n1+data.n2)
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k--
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if k <= 0 {
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break
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}
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idx := data.n2Idx
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idx++
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if idx >= len(nums2) {
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continue
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}
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heap.Push(pq, &pddata{
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n1: data.n1,
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n2: nums2[idx],
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n2Idx: idx,
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})
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}
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return res
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}
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type pddata struct {
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n1 int
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n2 int
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n2Idx int
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}
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type priorityQueue []*pddata
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func newPriorityQueue() *priorityQueue {
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pq := priorityQueue([]*pddata{})
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heap.Init(&pq)
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return &pq
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}
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func (pq priorityQueue) Len() int { return len(pq) }
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func (pq priorityQueue) Swap(i, j int) { pq[i], pq[j] = pq[j], pq[i] }
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func (pq priorityQueue) Less(i, j int) bool { return pq[i].n1+pq[i].n2 < pq[j].n1+pq[j].n2 }
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func (pq *priorityQueue) Pop() interface{} {
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old := *pq
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val := old[len(old)-1]
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old[len(old)-1] = nil
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*pq = old[0 : len(old)-1]
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return val
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}
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func (pq *priorityQueue) Push(i interface{}) {
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val := i.(*pddata)
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*pq = append(*pq, val)
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}
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package leetcode
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import (
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"fmt"
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"testing"
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)
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type question1439 struct {
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para1439
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ans1439
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}
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// para 是参数
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// one 代表第一个参数
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type para1439 struct {
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mat [][]int
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k int
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}
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// ans 是答案
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// one 代表第一个答案
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type ans1439 struct {
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one int
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}
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func Test_Problem1439(t *testing.T) {
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qs := []question1439{
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{
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para1439{[][]int{{1, 3, 11}, {2, 4, 6}}, 5},
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ans1439{7},
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},
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{
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para1439{[][]int{{1, 3, 11}, {2, 4, 6}}, 9},
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ans1439{17},
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},
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{
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para1439{[][]int{{1, 10, 10}, {1, 4, 5}, {2, 3, 6}}, 7},
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ans1439{9},
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},
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{
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para1439{[][]int{{1, 1, 10}, {2, 2, 9}}, 7},
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ans1439{12},
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},
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}
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fmt.Printf("------------------------Leetcode Problem 1439------------------------\n")
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for _, q := range qs {
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_, p := q.ans1439, q.para1439
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fmt.Printf("【input】:%v 【output】:%v\n", p, kthSmallest(p.mat, p.k))
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}
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fmt.Printf("\n\n\n")
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}
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@ -0,0 +1,144 @@
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# [1439. Find the Kth Smallest Sum of a Matrix With Sorted Rows](https://leetcode.com/problems/find-the-kth-smallest-sum-of-a-matrix-with-sorted-rows/)
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## 题目
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You are given an `m * n` matrix, `mat`, and an integer `k`, which has its rows sorted in non-decreasing order.
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You are allowed to choose exactly 1 element from each row to form an array. Return the Kth **smallest** array sum among all possible arrays.
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**Example 1:**
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```
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Input: mat = [[1,3,11],[2,4,6]], k = 5
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Output: 7
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Explanation: Choosing one element from each row, the first k smallest sum are:
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[1,2], [1,4], [3,2], [3,4], [1,6]. Where the 5th sum is 7.
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```
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**Example 2:**
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```
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Input: mat = [[1,3,11],[2,4,6]], k = 9
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Output: 17
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```
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**Example 3:**
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```
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Input: mat = [[1,10,10],[1,4,5],[2,3,6]], k = 7
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Output: 9
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Explanation: Choosing one element from each row, the first k smallest sum are:
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[1,1,2], [1,1,3], [1,4,2], [1,4,3], [1,1,6], [1,5,2], [1,5,3]. Where the 7th sum is 9.
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```
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**Example 4:**
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```
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Input: mat = [[1,1,10],[2,2,9]], k = 7
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Output: 12
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```
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**Constraints:**
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- `m == mat.length`
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- `n == mat.length[i]`
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- `1 <= m, n <= 40`
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- `1 <= k <= min(200, n ^ m)`
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- `1 <= mat[i][j] <= 5000`
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- `mat[i]` is a non decreasing array.
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## 题目大意
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给你一个 m * n 的矩阵 mat，以及一个整数 k ，矩阵中的每一行都以非递减的顺序排列。你可以从每一行中选出 1 个元素形成一个数组。返回所有可能数组中的第 k 个 最小数组和。
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## 解题思路
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- 这一题是第 373 题的升级版。在第 373 题中，给定 2 个有序数组，要求分别从这 2 个数组中选出一个数组成一个数对，最终输出和最小的 K 组。这一题中给出的是 m*n 的矩阵。其实是将第 373 题的 2 个数组升级为了 m 个数组。无非外层多了一层循环。这层循环依次从每一行中选出一个数，先从第 0 行和第 1 行取数，找到前 K 小的组合以后，再从第 2 行取数，以此类推。其他做法和第 373 题一致。维护一个长度为 k 的最小堆。每次从堆中 pop 出最小的数组和 sum 和对应的下标 index，然后依次将下标向后移动一位，生成新的 sum，加入堆中。
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## 代码
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```go
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package leetcode
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import "container/heap"
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func kthSmallest(mat [][]int, k int) int {
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if len(mat) == 0 || len(mat[0]) == 0 || k == 0 {
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return 0
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}
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prev := mat[0]
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for i := 1; i < len(mat); i++ {
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prev = kSmallestPairs(prev, mat[i], k)
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}
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if k < len(prev) {
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return -1
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}
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return prev[k-1]
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}
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func kSmallestPairs(nums1 []int, nums2 []int, k int) []int {
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res := []int{}
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if len(nums2) == 0 {
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return res
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}
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pq := newPriorityQueue()
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for i := 0; i < len(nums1) && i < k; i++ {
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heap.Push(pq, &pddata{
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n1: nums1[i],
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n2: nums2[0],
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n2Idx: 0,
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})
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}
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for pq.Len() > 0 {
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i := heap.Pop(pq)
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data := i.(*pddata)
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res = append(res, data.n1+data.n2)
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k--
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if k <= 0 {
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break
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}
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idx := data.n2Idx
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idx++
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if idx >= len(nums2) {
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continue
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}
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heap.Push(pq, &pddata{
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n1: data.n1,
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n2: nums2[idx],
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n2Idx: idx,
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})
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}
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return res
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}
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type pddata struct {
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n1 int
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n2 int
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n2Idx int
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}
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type priorityQueue []*pddata
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func newPriorityQueue() *priorityQueue {
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pq := priorityQueue([]*pddata{})
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heap.Init(&pq)
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return &pq
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}
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func (pq priorityQueue) Len() int { return len(pq) }
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func (pq priorityQueue) Swap(i, j int) { pq[i], pq[j] = pq[j], pq[i] }
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func (pq priorityQueue) Less(i, j int) bool { return pq[i].n1+pq[i].n2 < pq[j].n1+pq[j].n2 }
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func (pq *priorityQueue) Pop() interface{} {
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old := *pq
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val := old[len(old)-1]
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old[len(old)-1] = nil
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*pq = old[0 : len(old)-1]
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return val
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}
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func (pq *priorityQueue) Push(i interface{}) {
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val := i.(*pddata)
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*pq = append(*pq, val)
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}
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```
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