字符串全排列

剑指 Offer 38. 字符串的排列
字符串题目还是经常会遇到。

解法1：递归加去重

先固定一个位置，然后和其他位置的元素交换，依次往后推，当是最后一个时，是返回结果。

class Solution {
public:

    vector permutation(string s) {
        vector ret;
        sort(s.begin(), s.end());
        helper(ret, s, 0, s.size() - 1);
        return ret;
    }
    void helper(vector& ret, string& s, int left, int right) {
        if (left == right) {
            ret.push_back(s);
            return;
        }
        set st;
        for (int i = left; i <= right; ++ i) {
            if (st.find(s[i]) != st.end()) {
                continue;
            }
            st.insert(s[i]);
            swap(s[i], s[left]);
            helper(ret, s, left + 1, right);
            swap(s[i], s[left]);
        }
    }
};
[链接](https://leetcode-cn.com/problems/zi-fu-chuan-de-pai-lie-lcof/)

解法2：回溯

class Solution {
public:
    vector rec;
    vector vis;

    void backtrack(const string& s, int i, int n, string& perm) {
        if (i == n) {
            rec.push_back(perm);
            return;
        }
        for (int j = 0; j < n; j++) {
            if (vis[j] || (j > 0 && !vis[j - 1] && s[j - 1] == s[j])) {
                continue;
            }
            vis[j] = true;
            perm.push_back(s[j]);
            backtrack(s, i + 1, n, perm);
            perm.pop_back();
            vis[j] = false;
        }
    }

    vector permutation(string s) {
        int n = s.size();
        vis.resize(n);
        sort(s.begin(), s.end());
        string perm;
        backtrack(s, 0, n, perm);
        return rec;
    }
};

字典序排列

class Solution {
public:
    bool nextPermutation(string& s) {
        int i = s.size() - 2;
        while (i >= 0 && s[i] >= s[i + 1]) {
            i--;
        }
        if (i < 0) {
            return false;
        }
        int j = s.size() - 1;
        while (j >= 0 && s[i] >= s[j]) {
            j--;
        }
        swap(s[i], s[j]);
        reverse(s.begin() + i + 1, s.end());
        return true;
    }

    vector permutation(string s) {
        vector ret;
        sort(s.begin(), s.end());
        do {
            ret.push_back(s);
        } while (nextPermutation(s));
        return ret;
    }
};

Reference

[1] https://wizardforcel.gitbooks.io/the-art-of-programming-by-july/content/01.06.html