[LeetCode]31. Next Permutation
2016-04-12 14:53
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Problem Description
[]https://leetcode.com/problems/next-permutation/]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
思路
找规律的题,多写几个就能找到规律了。Code
package q031; public class Solution { public void nextPermutation(int[] nums) { int n = nums.length; if (n <= 1) return; int i; for (i = n - 1; i >= 1; i--) { if (nums[i] > nums[i - 1]) break; } if (i == 0) { reverse(nums, 0, n - 1); } else { int val = nums[i - 1]; int j = n - 1; while (j >= i) { if (nums[j] > val) break; j--; } swap(nums, j, i - 1); reverse(nums, i, n - 1); } } public void swap(int[] nums, int i, int j) { int tmp = nums[i]; nums[i] = nums[j]; nums[j] = tmp; // nums[i] = nums[i] ^ nums[j]; // nums[j] = nums[i] ^ nums[j]; // nums[i] = nums[i] ^ nums[j]; } public void reverse(int[] nums, int s, int e) { if (s > e) return; for (int i = s; i <= (s + e) / 2; i++) swap(nums, i, s + e - i); } }
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