31. Next Permutation(Medium)

版权声明：本文为博主原创文章，未经博主允许不得转载。 https://blog.csdn.net/NCUscienceZ/article/details/86923277

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 and use only constant 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

直接供上大佬的最优解法

class Solution {
    public void nextPermutation(int[] nums) {
      if(nums.length<=1){
          return;
      }
      
      int i= nums.length-1;
      
      for(;i>=1;i--){
         
         if(nums[i]>nums[i-1]){ //find first number which is smaller than it's after number
             break;
         }
      }
      
      if(i!=0){
          swap(nums,i-1); //if the number exist,which means that the nums not like{5,4,3,2,1}
      }
      
      reverse(nums,i);    
    }
    
    private void swap(int[] a,int i){
        for(int j = a.length-1;j>i;j--){
            if(a[j]>a[i]){
                int t = a[j];
                a[j] = a[i];
                a[i] = t;
                break;
            }
        }
    }
    
    private void reverse(int[] a,int i){//reverse the number after the number we have found
        int first = i;
        int last = a.length-1;
        while(first<last){
            int t = a[first];
            a[first] = a[last];
            a[last] = t;
            first ++;
            last --;
        }
    }
}