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	`1`	`+`
	`2`	`+# 0131.palindrome-partitioning`
	`3`	`+`
	`4`	+```text
	`5`	`+131. 分割回文串`
	`6`	`+`
	`7`	`+`
	`8`	`+给你一个字符串 s,请你将 s 分割成一些子串,使每个子串都是回文串。返回 s 所有可能的分割方案。`
	`9`	`+`
	`10`	`+回文串是正着读和反着读都一样的字符串。`
	`11`	`+`
	`12`	`+`
	`13`	`+`
	`14`	`+示例 1:`
	`15`	`+`
	`16`	`+输入:s = "aab"`
	`17`	`+输出:[["a","a","b"],["aa","b"]]`
	`18`	`+示例 2:`
	`19`	`+`
	`20`	`+输入:s = "a"`
	`21`	`+输出:[["a"]]`
	`22`	`+`
	`23`	`+`
	`24`	`+提示:`
	`25`	`+`
	`26`	`+1 <= s.length <= 16`
	`27`	`+s 仅由小写英文字母组成`
	`28`	`+`
	`29`	`+来源:力扣(LeetCode)`
	`30`	`+链接:https://leetcode-cn.com/problems/palindrome-partitioning`
	`31`	`+著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。`
	`32`	+```

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	`1`	`+package partitioning`
	`2`	`+`
	`3`	`+import "fmt"`
	`4`	`+`
	`5`	`+func Run() {`
	`6`	`+ s := "aab"`
	`7`	`+ p := partition(s)`
	`8`	`+ fmt.Println(p)`
	`9`	`+}`
	`10`	`+`
	`11`	`+func partition(s string) [][]string {`
	`12`	`+ n := len(s)`
	`13`	`+ f := make([][]bool, n)`
	`14`	`+ for i := range f {`
	`15`	`+ f[i] = make([]bool, n)`
	`16`	`+ for j := range f[i] {`
	`17`	`+ f[i][j] = true`
	`18`	`+ }`
	`19`	`+ }`
	`20`	`+ for i := n - 1; i >= 0; i-- {`
	`21`	`+ for j := i + 1; j < n; j++ {`
	`22`	`+ f[i][j] = s[i] == s[j] && f[i+1][j-1]`
	`23`	`+ }`
	`24`	`+ }`
	`25`	`+ ans := make([][]string, 0)`
	`26`	`+ splists := make([]string, 0)`
	`27`	`+ dfs := func(int) {}`
	`28`	`+ dfs = func(i int) {`
	`29`	`+ if i == n {`
	`30`	`+ ans = append(ans, append([]string{}, splists...))`
	`31`	`+ return`
	`32`	`+ }`
	`33`	`+ for j := i; j < n; j++ {`
	`34`	`+ if f[i][j] {`
	`35`	`+ splists = append(splists, s[i:j+1])`
	`36`	`+ dfs(j + 1)`
	`37`	`+ splists = splists[:len(splists)-1]`
	`38`	`+ }`
	`39`	`+ }`
	`40`	`+ }`
	`41`	`+ dfs(0)`
	`42`	`+ return ans`
	`43`	`+}`

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