[LeetCode] Next Permutation

Next Permutation

Implement next permutation, which rearranges numbers into the lexicographically next greater permutation of numbers.

If such arrangement is not possible, it must rearrange it as the 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 in the right-hand column.
1,2,3 → 1,3,2
3,2,1 → 1,2,3
1,1,5 → 1,5,1

解题思路：

这是一道算法题。学到了新的办法。开心。解法如下：

class Solution {
public:
    void nextPermutation(vector<int> &num) {
        int len = num.size();
        if(len<2){
            return;
        }
        //从右到左第一个逆序的
        int target1=-1;
        for(int i=len-2; i>=0; i--){
            if(num[i]<num[i+1]){
                target1=i;
                break;
            }
        }
        if(target1>=0){
            int target2=-1;
            for(int i=len-1; i>target1; i--){
                if(num[i]>num[target1]){
                    target2=i;
                    break;
                }
            }
            swap(num, target1, target2);
        }
        reverse(num, target1+1, len-1);
    }
    
    void swap(vector<int>& num, int i, int j){
        int temp=num[i];
        num[i]=num[j];
        num[j]=temp;
    }
    
    void reverse(vector<int>& num, int i, int j){
        while(i<j){
            swap(num, i, j);
            i++;
            j--;
        }
    }
};

[LeetCode] Next Permutation

原文：http://blog.csdn.net/kangrydotnet/article/details/45343569