剑指offer 字符串的排列

class Solution {public:vector<string> Permutation(string str) {sort(str.begin(), str.end());string tem = str;vector<string> ans;while (tem != "") {ans.push_back(tem);getnext(tem);}return ans;}void getnext(string &str) {int len = str.size();int i;int judge = 0;int cur = len - 2;while (cur >= 0) {for (i = len - 1; i > cur; i--) {if (str[i] > str[cur]) {judge = 1;break;}}if (judge) {swap(str[i], str[cur]);reverse(str.begin() + cur + 1, str.end());break;}cur--;}if (!judge)str = "";}};

 

运用了求一个排列的下一个排列算法，可以用递归，更加简单。