131. 分割回文串

原题

https://leetcode.cn/problems/palindrome-partitioning/description/

思路

深度优先搜索

复杂度

时间：O(n)
空间：O(n)

Python代码

class Solution:
    def partition(self, s: str) -> List[List[str]]:
        ans = []

        def dfs(start, path):
            if start == len(s):
                ans.append(path[:])
                return
            for end in range(start, len(s)):
                sub = s[start : end + 1]
                if sub == sub[::-1]:
                    path.append(sub)
                    dfs(end + 1, path)
                    # 还原
                    path.pop()

        dfs(0, [])
        return ans

Go代码

func reverse(s string) string {
	slice := []rune(s)
	for i, j := 0, len(s)-1; i < j; i, j = i+1, j-1 {
		slice[i], slice[j] = slice[j], slice[i]
	}
	return string(slice)
}

func partition(s string) [][]string {
	var ans [][]string
	var dfs func(int, []string)
	dfs = func(start int, path []string) {
		if start == len(s) {
			pathCopy := make([]string, len(path))
			copy(pathCopy, path)
			ans = append(ans, pathCopy)
			return
		}
		for end := start; end < len(s); end++ {
			sub := s[start : end+1]
			if sub == reverse(sub) {
				path = append(path, sub)
				dfs(end+1, path)
				// 还原
				path = path[:len(path)-1]
			}
		}
	}

	dfs(0, []string{})
	return ans
}