leetcode-全排列（permutation）

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给定一个数字列表，返回其所有可能的排列。

class Solution {public:    /**     * @param nums: A list of integers.     * @return: A list of permutations.     */    vector<vector<int> > permute(vector<int> nums) {        // write your code here        int len = nums.size();        vector<vector<int> > res;        if (len == 0) {            res.push_back(nums);            return res;        }        sort(nums.begin(), nums.end());        permutation(res, nums, len-1);        return res;    }    void permutation(vector<vector<int>>& res, vector<int>& nums, int n) {        if (n == 0) {            res.push_back(nums);            return;        }        for (int i = 0; i <= n; i++) {            swap(nums[i], nums[n]);            permutation(res, nums, n - 1);            swap(nums[i], nums[n]);        }    }};

非递归版本

// version 1: Non-Recursionclass Solution {public:    /**     * @param nums: A list of integers.     * @return: A list of permutations.     */    vector<vector<int> > permute(vector<int> nums) {        vector<vector<int> > permutations;        if (nums.size() == 0) {            permutations.push_back(vector<int>());            return permutations;        }        int n = nums.size();        vector<int> stack;        bool inStack[n];        for (int i = 0; i < n; i++) {            inStack[i] = false;        }        stack.push_back(-1);        while (stack.size() != 0) {            // pop the last             int last = stack[stack.size() - 1];            stack.pop_back();            if (last != -1) {                inStack[last] = false;            }            // increase the last, find the next bigger & avaiable number            int next = -1;            for (int i = last + 1; i < n; i++) {                if (inStack[i] == false) {                    next = i;                    break;                }            }            if (next == -1) {                continue;            }            // generate the next permutation            stack.push_back(next);            inStack[next] = true;            for (int i = 0; i < n; i++) {                if (!inStack[i]) {                    stack.push_back(i);                    inStack[i] = true;                }            }            // generate real permutation from index            vector<int> permutation;            for (int i = 0; i < n; i++) {                permutation.push_back(nums[stack[i]]);            }            permutations.push_back(permutation);        }        return permutations;    }};

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