leetcode : Next Permutation

问题描述：

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

原问题链接：https://leetcode.com/problems/next-permutation/

问题分析

 关于这个问题的分析在之前的一篇文章里分析过。这里就不再详述。针对这个问题我们只是需要稍微做一点修改。假设最开始的串是一个纯递减的。也就是说它达到一个串最大的值了。那么只需要将它翻转一下就可以了。详细的代码实现如下：

public class Solution {
    public void nextPermutation(int[] num) {
        int lastInc = findLastIncr(num);
        if(lastInc < 0)
            reverse(num, 0, num.length - 1);
        else {
            int l = findLastBigger(num, lastInc);
            swap(num, lastInc, l);
            reverse(num, lastInc + 1, num.length - 1);
        }
    }
    
    private int findLastIncr(int[] num) {
        for(int i = num.length - 1; i >= 1; i--) {
            if(num[i] > num[i - 1])
                return i - 1;
        }
        
        return -1;
    }
    
    private int findLastBigger(int[] num, int lastIdx) {
        for(int i = num.length - 1; i > lastIdx; i--)
            if(num[i] > num[lastIdx])
                return i;
        return -1;
    }
    
    private void reverse(int[] num, int l, int r) {
        while(l < r) {
            swap(num, l, r);
            l++;
            r--;
        }
    }
    
    private void swap(int[] num, int i, int j) {
        int temp = num[i];
        num[i] = num[j];
        num[j] = temp;
    }
}