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https://github.com/halfrost/LeetCode-Go.git
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规范格式
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package leetcode
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import "sort"
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// 解法一 桶排序,时间复杂度 O(n^2)
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func relativeSortArray(A, B []int) []int {
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count := [1001]int{}
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for _, a := range A {
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count[a]++
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}
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res := make([]int, 0, len(A))
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for _, b := range B {
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for count[b] > 0 {
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res = append(res, b)
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count[b]--
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}
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}
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for i := 0; i < 1001; i++ {
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for count[i] > 0 {
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res = append(res, i)
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count[i]--
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}
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}
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return res
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}
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// 解法二 模拟,时间复杂度 O(n^2)
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func relativeSortArray1(arr1 []int, arr2 []int) []int {
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leftover, m, res := []int{}, make(map[int]int), []int{}
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for _, v := range arr1 {
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m[v]++
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}
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for _, s := range arr2 {
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count := m[s]
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for i := 0; i < count; i++ {
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res = append(res, s)
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}
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m[s] = 0
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}
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for v, count := range m {
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for i := 0; i < count; i++ {
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leftover = append(leftover, v)
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}
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}
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sort.Ints(leftover)
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res = append(res, leftover...)
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return res
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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 question1122 struct {
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para1122
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ans1122
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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 para1122 struct {
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arr1 []int
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arr2 []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 ans1122 struct {
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one []int
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}
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func Test_Problem1122(t *testing.T) {
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qs := []question1122{
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question1122{
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para1122{[]int{2, 3, 1, 3, 2, 4, 6, 7, 9, 2, 19}, []int{2, 1, 4, 3, 9, 6}},
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ans1122{[]int{2, 2, 2, 1, 4, 3, 3, 9, 6, 7, 19}},
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},
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}
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fmt.Printf("------------------------Leetcode Problem 1122------------------------\n")
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for _, q := range qs {
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_, p := q.ans1122, q.para1122
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fmt.Printf("【input】:%v 【output】:%v\n", p, relativeSortArray(p.arr1, p.arr2))
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}
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fmt.Printf("\n\n\n")
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}
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45
leetcode/1122.Relative-Sort-Array/README.md
Executable file
45
leetcode/1122.Relative-Sort-Array/README.md
Executable file
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# [1122. Relative Sort Array](https://leetcode.com/problems/relative-sort-array/)
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## 题目:
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Given two arrays `arr1` and `arr2`, the elements of `arr2` are distinct, and all elements in `arr2` are also in `arr1`.
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Sort the elements of `arr1` such that the relative ordering of items in `arr1` are the same as in `arr2`. Elements that don't appear in `arr2` should be placed at the end of `arr1` in **ascending** order.
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**Example 1:**
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Input: arr1 = [2,3,1,3,2,4,6,7,9,2,19], arr2 = [2,1,4,3,9,6]
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Output: [2,2,2,1,4,3,3,9,6,7,19]
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**Constraints:**
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- `arr1.length, arr2.length <= 1000`
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- `0 <= arr1[i], arr2[i] <= 1000`
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- Each `arr2[i]` is distinct.
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- Each `arr2[i]` is in `arr1`.
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## 题目大意
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给你两个数组,arr1 和 arr2,
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- arr2 中的元素各不相同
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- arr2 中的每个元素都出现在 arr1 中
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对 arr1 中的元素进行排序,使 arr1 中项的相对顺序和 arr2 中的相对顺序相同。未在 arr2 中出现过的元素需要按照升序放在 arr1 的末尾。
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提示:
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- arr1.length, arr2.length <= 1000
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- 0 <= arr1[i], arr2[i] <= 1000
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- arr2 中的元素 arr2[i] 各不相同
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- arr2 中的每个元素 arr2[i] 都出现在 arr1 中
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## 解题思路
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- 给出 2 个数组 A 和 B,A 中包含 B 中的所有元素。要求 A 按照 B 的元素顺序排序,B 中没有的元素再接着排在后面,从小到大排序。
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- 这一题有多种解法。一种暴力的解法就是依照题意,先把 A 中的元素都统计频次放在 map 中,然后 按照 B 的顺序输出,接着再把剩下的元素排序接在后面。还有一种桶排序的思想,由于题目限定了 A 的大小是 1000,这个数量级很小,所以可以用 1001 个桶装所有的数,把数都放在桶里,这样默认就已经排好序了。接下来的做法和前面暴力解法差不多,按照频次输出。B 中以外的元素就按照桶的顺序依次输出即可。
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