LeetCode | # 31. Next Permutation-优快云博客

本文链接：https://blog.youkuaiyun.com/Karmen199x/article/details/102826907

本文详细解析了NextPermutation算法，介绍了如何实现数组元素的字典序下一个排列。通过从右往左查找逆序位置，交换元素并反转子数组，实现了在原地使用常数额外内存完成任务。

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31. Next Permutation

Description

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

Solution: (Java)

class Solution {
    public void nextPermutation(int[] nums) {
        int swap_index = -1, temp;
        for (int i = nums.length-2; i >= 0; i--) {
            if (nums[i] < nums[i+1]) {
                // 从右往左，找到第一个逆序的位置
                swap_index = i;
                for (int j = nums.length-1; j > swap_index; j--) {
                    if (nums[j] > nums[swap_index]) {
                        // 从右往左(也可以从左往右，找到最后一个比逆序位置数大的)，找到第一个比逆序位置数大的，交换两者
                        temp = nums[swap_index];
                        nums[swap_index] = nums[j];
                        nums[j] = temp;
                        break;
                    }
                }
                break;
            }
        }
        int left = swap_index+1;
        int right = nums.length-1;
        // 将后面的降序的数组翻转为升序
        while (left < right) {
            temp = nums[left];
            nums[left] = nums[right];
            nums[right] = temp;
            left++;
            right--;
        }
    }
}