leetcode[31]Next Permutation

最新推荐文章于 2020-07-31 23:36:22 发布

转载最新推荐文章于 2020-07-31 23:36:22 发布 · 113 阅读

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原文链接：http://www.cnblogs.com/Vae1990Silence/p/4283610.html

本文介绍了一个用于重新排列数字数组以获得下一个更大排列的算法。如果不存在更大的排列，则将其排列为最小可能的顺序（即升序）。该算法在原地进行操作，不使用额外内存。

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

class Solution {
public:
    void nextPermutation(vector<int> &num) {
    int i=0,j=0,k=0;
    int n=num.size();
    if (n==0||n==1)return;
    if(n==2)
    {
       reverse(num.begin(),num.end());
       return;
    }
    int jj=0;
    for(;jj<num.size()-1;jj++)
    {
        if (num[jj]<num[jj+1])
        {
            break;
        }
    }
    if(jj==num.size()-1)
    {
        reverse(num.begin(),num.end());
        return;
    }
    for (j=num.size()-2;j>=0;j--)
    {
        if(num[j]<num[j+1])break;
    }
    i=j+1;
    for (k=num.size()-1;k>=0;k--)
    {
        if(num[i-1]<num[k])break;
    }
    j=k;
    int temp=num[i-1];
    num[i-1]=num[j];
    num[j]=temp;
    reverse(num.begin()+i,num.end());     
    }
};

转载于:https://www.cnblogs.com/Vae1990Silence/p/4283610.html