leetcode31. Next Permutation

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Next Permutation

参考：https://yq.aliyun.com/articles/863

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

解题思路：把这个题想成是求数组中的元素所组成的数中比比当前数大的一个最小数，这个就容易理解了，从后后往前找，找到第一个递增的位置i<j,在j以后的数列全部是递减的，找到比i大的数中最小的与i互换位置，然后将j以后的数列变为递增。此时就得到结果了。

package leetcode;import java.util.Arrays;public class leet31 {public static void main(String[] args) {int[] nums = { 3, 2, 1 };leet31 leet = new leet31();leet.nextPermutation(nums);}   public void nextPermutation(int[] nums) {int i;for (i = nums.length - 1; i > 0; i--) {//从后往前遍历，找到递增位置iif (nums[i] > nums[i - 1]) {for (int j = i; j < nums.length; j++) {//从递增的位置往后找到比i大的数中的最小数if (j == nums.length - 1) {//数列全部递增时int tmp = nums[i - 1];nums[i - 1] = nums[j];nums[j] = tmp;} else if (nums[j] > nums[i - 1] && nums[j + 1] < nums[i - 1]) {int tmp = nums[i - 1];nums[i - 1] = nums[j];nums[j] = tmp;for (int low = i - 1 + 1, high = nums.length - 1; low < high; low++, high--) {//i位置后的递减序列变递增int temp = nums[low];nums[low] = nums[high];nums[high] = temp;}}}System.out.println(Arrays.toString(nums));return;}}if (i == 0) {Arrays.sort(nums);System.out.println(Arrays.toString(nums));}}}

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