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
代码
Approach #1
Find first decreasing element i
Find number nums[j] greater than nums[i]
swap
reverse i + 1 之后的所有元素
classSolution {publicvoidnextPermutation(int[] nums) {// 1. find the first decreasing elementint i =nums.length-2;while (i >=0&& nums[i] >= nums[i +1]) { i--; }// 2. find an element greather than nums[i]if (i >=0) {int j =nums.length-1;while (j >=0&& nums[j] <= nums[i]) { j--; }// 3. swap nums[i] and nums[j]swp(nums, i, j); }// 4. reverse element from i + 1reverse(nums, i +1); }privatevoidreverse(int[] nums,int start) {int i = start, j =nums.length-1;while (i < j) {swap(nums, i, j); i++; j--; } }privatevoidswap(int[] nums,int i,int j) {int temp = nums[i]; nums[i] = nums[j]; nums[j] = temp; }}