Permutations

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Given a collection of numbers, return all possible permutations.

For example,
[1,2,3] have the following permutations:

[1,2,3], [1,3,2], [2,1,3], [2,3,1], [3,1,2], and [3,2,1].

首先排序

1.先从右到左，找到第一个违反递减规律的数num[i]

2.再从右到i+1 找到第一个比num[i]大的数字num[j]

3.交换num[i] 与num[j]

4.对i+1到num.end() 排序

swap+sort 策略

class Solution {public:    bool next(vector<int>& num){        int i=num.size()-2,j=num.size()-1;        while(i>=0&&num[i]>=num[i+1]) i--;        if(i<0) return false;        while(j>i&&num[j]<=num[i]) j--;        swap(num[i],num[j]);        if(i+1<num.size()-1)            sort(num.begin()+i+1,num.end());        return true;    }    vector<vector<int> > permute(vector<int> &num) {        sort(num.begin(),num.end());        vector<vector<int>> result;        do{            result.push_back(num);        }while(next(num));        return result;    }};

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