LeetCode – Next Permutation (Java)
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 inplace, do not allocate extra memory.
Here are some examples. Inputs are in the lefthand column and its corresponding outputs are in the righthand column.
1,2,3 → 1,3,2 3,2,1 → 1,2,3 1,1,5 → 1,5,1
Analysis
The steps to solve this problem:
1) scan from right to left, find the first element that is less than its previous one.
4 5 6 3 2 1  p
2) scan from right to left, find the first element that is greater than p.
4 5 6 3 2 1  q
3) swap p and q
4 5 6 3 2 1 swap 4 6 5 3 2 1
4) reverse elements [p+1, nums.length]
4 6 1 2 3 5
Java Solution
public void nextPermutation(int[] nums) { if(nums == null  nums.length<2) return; int p=0; for(int i=nums.length2; i>=0; i){ if(nums[i]<nums[i+1]){ p=i; break; } } int q = 0; for(int i=nums.length1; i>p; i){ if(nums[i]> nums[p]){ q=i; break; } } if(p==0 && q==0){ reverse(nums, 0, nums.length1); return; } int temp=nums[p]; nums[p]=nums[q]; nums[q]=temp; if(p<nums.length1){ reverse(nums, p+1, nums.length1); } } public void reverse(int[] nums, int left, int right){ while(left<right){ int temp = nums[left]; nums[left]=nums[right]; nums[right]=temp; left++; right; } } 
<pre><code> String foo = "bar"; </code></pre>

vishwas s

A

Kyle