31. Next Permutation.md

31. Next Permutation

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leetcode-cn Daily Challenge on November 10th, 2020.

leetcode Daily Challenge on January 31th, 2021.

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Difficulty : Medium

Related Topics : Array

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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

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Solution

• mine
  • Java
    • Runtime: 0 ms, faster than 100.00%, Memory Usage: 39.4 MB, less than 50.00% of Java online submissions

      // O(N)time 
      // O(1)space
      public void nextPermutation(int[] nums) {
          int len = nums.length;
          if (len < 2) {
              return;
          }
          int index = len - 1;
          while (index > 0 && nums[index - 1] >= nums[index]) {
              index--;
          }
          if (index != 0) {
              int val = nums[index - 1];
              int j = len - 1;
              //find the last i which make nums[i] > val
              while (j >= index) {
                  if (nums[j] > val)
                      break;
                  j--;
              }
              int t = nums[j];
              nums[j] = nums[index - 1];
              nums[index - 1] = t;
          }
      
          int s = index, e = len - 1;
          while (s < e) {
              int t = nums[s];
              nums[s] = nums[e];
              nums[e] = t;
              s++;
              e--;
          }
      }
      

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• the most votes

• Runtime: 0 ms, faster than 100.00%, Memory Usage: 39.4 MB, less than 50.00% of Java online submissions

  public void nextPermutation(int[] num) {
      int n=num.length;
      if(n<2)
          return;
      int index=n-1;
      while(index>0){
          if(num[index-1]<num[index])
              break;
          index--;
      }
      if(index==0){
          reverseSort(num,0,n-1);
          return;
      }
      else{
          int val=num[index-1];
          int j=n-1;
          while(j>=index){
              if(num[j]>val)
                  break;
              j--;
          }
          swap(num,j,index-1);
          reverseSort(num,index,n-1);
          return;
      }
  }
  
  public void swap(int[] num, int i, int j){
      int temp=0;
      temp=num[i];
      num[i]=num[j];
      num[j]=temp;
  }
  
  public void reverseSort(int[] num, int start, int end){
      if(start>end)
          return;
      for(int i=start;i<=(end+start)/2;i++)
          swap(num,i,start+end-i);
  }
  

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