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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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const primeRK = 16777619
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// 解法一 二分搜索 + Rabin-Karp
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func findLength(A []int, B []int) int {
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low, high := 0, min(len(A), len(B))
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for low < high {
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mid := (low + high + 1) >> 1
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if hasRepeated(A, B, mid) {
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low = mid
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} else {
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high = mid - 1
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}
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}
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return low
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}
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func min(a int, b int) int {
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if a > b {
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return b
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}
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return a
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}
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func hashSlice(arr []int, length int) []int {
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// hash 数组里面记录 arr 比 length 长出去部分的 hash 值
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hash, pl, h := make([]int, len(arr)-length+1), 1, 0
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for i := 0; i < length-1; i++ {
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pl *= primeRK
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}
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for i, v := range arr {
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h = h*primeRK + v
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if i >= length-1 {
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hash[i-length+1] = h
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h -= pl * arr[i-length+1]
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}
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}
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return hash
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}
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func hasSamePrefix(A, B []int, length int) bool {
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for i := 0; i < length; i++ {
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if A[i] != B[i] {
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return false
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}
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}
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return true
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}
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func hasRepeated(A, B []int, length int) bool {
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hs := hashSlice(A, length)
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hashToOffset := make(map[int][]int, len(hs))
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for i, h := range hs {
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hashToOffset[h] = append(hashToOffset[h], i)
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}
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for i, h := range hashSlice(B, length) {
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if offsets, ok := hashToOffset[h]; ok {
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for _, offset := range offsets {
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if hasSamePrefix(A[offset:], B[i:], length) {
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return true
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}
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}
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}
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}
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return false
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}
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// 解法二 DP 动态规划
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func findLength1(A []int, B []int) int {
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res, dp := 0, make([][]int, len(A)+1)
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for i := range dp {
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dp[i] = make([]int, len(B)+1)
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}
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for i := len(A) - 1; i >= 0; i-- {
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for j := len(B) - 1; j >= 0; j-- {
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if A[i] == B[j] {
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dp[i][j] = dp[i+1][j+1] + 1
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if dp[i][j] > res {
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res = dp[i][j]
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}
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}
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}
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}
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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 question718 struct {
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para718
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ans718
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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 para718 struct {
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A []int
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B []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 ans718 struct {
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one int
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}
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func Test_Problem718(t *testing.T) {
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qs := []question718{
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question718{
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para718{[]int{0, 0, 0, 0, 0}, []int{0, 0, 0, 0, 0}},
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ans718{5},
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},
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question718{
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para718{[]int{1, 2, 3, 2, 1}, []int{3, 2, 1, 4, 7}},
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ans718{3},
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},
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question718{
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para718{[]int{0, 0, 0, 0, 1}, []int{1, 0, 0, 0, 0}},
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ans718{4},
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},
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question718{
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para718{[]int{0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}, []int{0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0}},
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ans718{59},
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},
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}
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fmt.Printf("------------------------Leetcode Problem 718------------------------\n")
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for _, q := range qs {
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_, p := q.ans718, q.para718
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fmt.Printf("【input】:%v 【output】:%v\n", p, findLength(p.A, p.B))
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}
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fmt.Printf("\n\n\n")
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}
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37
leetcode/0718.Maximum-Length-of-Repeated-Subarray/README.md
Executable file
37
leetcode/0718.Maximum-Length-of-Repeated-Subarray/README.md
Executable file
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# [718. Maximum Length of Repeated Subarray](https://leetcode.com/problems/maximum-length-of-repeated-subarray/)
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## 题目:
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Given two integer arrays `A` and `B`, return the maximum length of an subarray that appears in both arrays.
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**Example 1:**
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Input:
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A: [1,2,3,2,1]
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B: [3,2,1,4,7]
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Output: 3
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Explanation:
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The repeated subarray with maximum length is [3, 2, 1].
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**Note:**
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1. 1 <= len(A), len(B) <= 1000
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2. 0 <= A[i], B[i] < 100
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## 题目大意
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给两个整数数组 A 和 B ，返回两个数组中公共的、长度最长的子数组的长度。
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## 解题思路
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- 给出两个数组，求这两个数组中最长相同子串的长度。
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- 这一题最容易想到的是 DP 动态规划的解法。`dp[i][j]` 代表在 A 数组中以 `i` 下标开始的子串与 B 数组中以 `j` 下标开始的子串最长相同子串的长度，状态转移方程为 `dp[i][j] = dp[i+1][j+1] + 1` (当 `A[i] == B[j]`)。这种解法的时间复杂度是 O(n^2)，空间复杂度 O(n^2)。
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- 这一题最佳解法是二分搜索 + `Rabin-Karp`。比较相同子串耗时的地方在于，需要一层循环，遍历子串所有字符。但是如果比较两个数字就很快，`O(1)` 的时间复杂度。所以有人就想到了，能不能把字符串也映射成数字呢？这样比较起来就非常快。这个算法就是 `Rabin-Karp` 算法。字符串映射成一个数字不能随意映射，还要求能根据字符串前缀动态增加，比较下一个字符串的时候，可以利用已比较过的前缀，加速之后的字符串比较。在 Rabin-Karp 算法中有一个“码点”的概念。类似于10进制中的进制。具体的算法讲解可以见这篇：
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[基础知识 - Rabin-Karp 算法](https://www.cnblogs.com/golove/p/3234673.html)
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“码点”一般取值为一个素数。在 go 的 `strings` 包里面取值是 16777619。所以这一题也可以直接取这个值。由于这一次要求我们找最长长度，所以把最长长度作为二分搜索的目标。先将数组 A 和数组 B 中的数字都按照二分出来的长度，进行 `Rabin-Karp` hash。对 A 中的 hash 与下标做映射关系，存到 map 中，方便后面快速查找。然后遍历 B 中的 hash，当 hash 一致的时候，再匹配下标。如果下标存在，且拥有相同的前缀，那么就算找到了相同的子串了。最后就是不断的二分，找到最长的结果即可。这个解法的时间复杂度 O(n * log n)，空间复杂度 O(n)。
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