Implement next permutation, which rearranges numbers into the lexicographically next greater permutation of numbers.If such arrangement is lowest possible order(ie, sorted in ascending order).The replacement must be in-place, do not allocate extra memory.Here are some examples. Inputs are in the left-hand column and its corresponding outputs are -hand column.,1,5 → 1,5,1
基本思想：
这儿由于数组是反向遍历的，所以可以使用使用stl中的反向迭代器。 runtime:12ms
class Solution {public:void nextPermutation(vector<int>& nums) {if(nums.size()<2) return ;auto iter=nums.rbegin();while(iter!=(nums.rend()-1)&&*iter<=*(iter+1))iter++;if(iter==nums.rend()-1)sort(nums.begin(),nums.end());else{auto upper=iter;auto tmp=nums.rbegin();for(;tmp!=iter;tmp++){if(*tmp>*(iter+1)){if(*tmp<*upper){upper=tmp;}}}swap(*(iter+1),*upper);sort(nums.rbegin(),iter+1,greater<int>());}}};
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