[Vssue]0131.分割回文串.md

[youngyangyang04]

https://www.programmercarl.com/0131.%E5%88%86%E5%89%B2%E5%9B%9E%E6%96%87%E4%B8%B2.html

[Du1in9]

private void backtracking(String s, int start) {
    if (start >= s.length()) {
        paths.add(new ArrayList<>(path));
        return;
    }
    for (int i = start; i < s.length(); i++) {
        if (isPalindrome(s, start, i)) {
            path.add(s.substring(start, i + 1));
            backtracking(s, i + 1);
            path.remove(path.size() - 1);
        }
    }
}

// 深度1: backtracking("aab", 0), path = ""
i = 0："a" 是回文串, path = ["a"]，递归 backtracking("aab", 1)，回溯，path = []
i = 1："aa" 是回文串, path = ["aa"]，递归 backtracking("aab", 2)，回溯，path = []
i = 2："aab" 不是回文串, 停止遍历（剪枝）
// 深度2: 例: backtracking("aab", 1), path = ["a"]
i = 1："a" 是回文串, path = ["a","a"]，递归 backtracking("aab", 2)，回溯，path = ["a"]
i = 2："ab" 不是回文串, 停止遍历（剪枝）
// 深度3: 例: backtracking("aab", 2), path = ["a","a"]
i = 2："b" 是回文串, path = ["a","a","b"]，递归 3 >= 3，加入并返回。回溯，path = ["a","a"]