解题思路

首先需要一个check函数用来验证字符串是否为回文串，这里check我们选用双指针。

从下标0开始不断查找回文串，当下标到达s长度时添加进result

代码

Python

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

        def check(s: str):
            l, r = 0, len(s) - 1
            while l <= r:
                if s[l] != s[r]: return False
                l += 1
                r -= 1
            return True

        def backtracking(temp: list[str], start: int):
            if start == len(s):
                result.append(temp.copy())
                return
            sub = ""
            for i in range(start, len(s)):
                sub += s[i]
                if check(sub):
                    temp.append(sub)
                    backtracking(temp, i + 1)
                    temp.pop()

        backtracking([], 0)
        return result

C++

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class Solution {
private:
    vector<vector<int>> f;
    vector<vector<string>> ret;
    vector<string> ans;
    int n;

public:
    void dfs(const string& s, int i) {
        if (i == n) {
            ret.push_back(ans);
            return;
        }
        for (int j = i; j < n; ++j) {
            if (f[i][j]) {
                ans.push_back(s.substr(i, j - i + 1));
                dfs(s, j + 1);
                ans.pop_back();
            }
        }
    }

    vector<vector<string>> partition(string s) {
        n = s.size();
        f.assign(n, vector<int>(n, true));

        for (int i = n - 1; i >= 0; --i) {
            for (int j = i + 1; j < n; ++j) {
                f[i][j] = (s[i] == s[j]) && f[i + 1][j - 1];
            }
        }

        dfs(s, 0);
        return ret;
    }
};

Go

func partition(s string) (ans [][]string) {
    n := len(s)
    f := make([][]int8, n)
    for i := range f {
        f[i] = make([]int8, n)
    }

    // 0 表示尚未搜索，1 表示是回文串，-1 表示不是回文串
    var isPalindrome func(i, j int) int8
    isPalindrome = func(i, j int) int8 {
        if i >= j {
            return 1
        }
        if f[i][j] != 0 {
            return f[i][j]
        }
        f[i][j] = -1
        if s[i] == s[j] {
            f[i][j] = isPalindrome(i+1, j-1)
        }
        return f[i][j]
    }

    splits := []string{}
    var dfs func(int)
    dfs = func(i int) {
        if i == n {
            ans = append(ans, append([]string(nil), splits...))
            return
        }
        for j := i; j < n; j++ {
            if isPalindrome(i, j) > 0 {
                splits = append(splits, s[i:j+1])
                dfs(j + 1)
                splits = splits[:len(splits)-1]
            }
        }
    }
    dfs(0)
    return
}