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Commit e37fe67

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Merge pull request youngyangyang04#2643 from markwang1992/131-partition
131.分割回文串增加Go动态规划优化回文串解法
2 parents 37af407 + 11ddcbd commit e37fe67

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‎problems/0093.复原IP地址.md‎

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@@ -143,7 +143,7 @@ for (int i = startIndex; i < s.size(); i++) {
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代码如下:
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```CPP
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// 判断字符串s在左闭又闭区间[start, end]所组成的数字是否合法
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// 判断字符串s在左闭右闭区间[start, end]所组成的数字是否合法
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bool isValid(const string& s, int start, int end) {
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if (start > end) {
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return false;
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} else break; // 不合法,直接结束本层循环
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}
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}
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// 判断字符串s在左闭又闭区间[start, end]所组成的数字是否合法
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// 判断字符串s在左闭右闭区间[start, end]所组成的数字是否合法
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bool isValid(const string& s, int start, int end) {
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if (start > end) {
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return false;

‎problems/0131.分割回文串.md‎

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@@ -527,6 +527,7 @@ class Solution:
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```
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### Go
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回溯 基本版
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```go
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var (
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path []string // 放已经回文的子串
@@ -565,6 +566,63 @@ func isPalindrome(s string) bool {
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}
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```
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回溯+动态规划优化回文串判断
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```go
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var (
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result [][]string
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path []string // 放已经回文的子串
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isPalindrome [][]bool // 放事先计算好的是否回文子串的结果
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)
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func partition(s string) [][]string {
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result = make([][]string, 0)
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path = make([]string, 0)
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computePalindrome(s)
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backtracing(s, 0)
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return result
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}
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func backtracing(s string, startIndex int) {
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// 如果起始位置已经大于s的大小,说明已经找到了一组分割方案了
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if startIndex >= len(s) {
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tmp := make([]string, len(path))
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copy(tmp, path)
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result = append(result, tmp)
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return
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}
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for i := startIndex; i < len(s); i++ {
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if isPalindrome[startIndex][i] { // 是回文子串
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// 获取[startIndex,i]在s中的子串
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path = append(path, s[startIndex:i+1])
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} else { // 不是回文,跳过
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continue
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}
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backtracing(s, i + 1) // 寻找i+1为起始位置的子串
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path = path[:len(path)-1] // 回溯过程,弹出本次已经添加的子串
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}
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}
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func computePalindrome(s string) {
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// isPalindrome[i][j] 代表 s[i:j](双边包括)是否是回文字串
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isPalindrome = make([][]bool, len(s))
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for i := 0; i < len(isPalindrome); i++ {
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isPalindrome[i] = make([]bool, len(s))
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}
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for i := len(s)-1; i >= 0; i-- {
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// 需要倒序计算, 保证在i行时, i+1行已经计算好了
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for j := i; j < len(s); j++ {
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if j == i {
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isPalindrome[i][j] = true
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} else if j - i == 1 {
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isPalindrome[i][j] = s[i] == s[j]
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} else {
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isPalindrome[i][j] = s[i] == s[j] && isPalindrome[i+1][j-1]
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}
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}
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}
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}
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```
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### JavaScript
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```js

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