【题目】

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

【分析】

（1）From right to left,find first digit which violate the increase  call it partitionNumber

（2）partitionNumber = -1 indicate largest ,should （3）,if not From right to left,find first digit which large than partitionNumber call it changeNumber   

swap the partition index and changeNumber

（3）reverse all the digit on the right of partition index

【代码】

/*********************************
*   日期：2015-01-16
*   作者：SJF0115
*   题目: 31.Next Permutation
*   网址：https://oj.leetcode.com/problems/next-permutation/
*   结果：AC
*   来源：LeetCode
*   博客：
**********************************/
#include <iostream>
#include <vector>
using namespace std;

class Solution {
public:
    void nextPermutation(vector<int> &num) {
        int count = num.size();
        if(count == 0 || count == 1){
            return;
        }//if
        // From right to left,find first digit which violate the increase
        // call it partitionNumber
        int index;
        for(index = count-2;index >= 0;index--){
            if(num[index+1] > num[index]){
                break;
            }//if
        }//for
        int tmp;
        // -1 indicate largest
        if(index != -1){
            // From right to left,find first digit which large than partitionNumber
            // call it changeNumber
            for(int i = count-1;i > index;i--){
                if(num[index] < num[i]){
                    // swap the partition index and changeNumber
                    tmp = num[index];
                    num[index] = num[i];
                    num[i] = tmp;
                    break;
                }//if
            }//for
        }//if
        // reverse all the digit on the right of partition index
        for(int i = index+1,j = count - 1;i < j;i++,j--){
            tmp = num[i];
            num[i] = num[j];
            num[j] = tmp;
        }//for
    }
};

int main(){
    Solution solution;
    vector<int> num;
    num.push_back(1);
    num.push_back(1);
    num.push_back(5);
    // 重新排列
    solution.nextPermutation(num);
    // 输出
    for(int i = 0;i < num.size();i++){
        cout<<num[i];
    }
    cout<<endl;
    return 0;
}

[LeetCode]31.Next Permutation