31. Next Permutation

最新推荐文章于 2020-12-07 13:42:30 发布

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本文介绍了一个算法，用于在原地将数组中的整数序列重新排列为字典序中下一个更大的排列。如果这样的排列不存在，则将其变为最小可能的顺序（即升序）。文章详细解释了寻找并替换元素的过程，并给出了具体的实现代码。

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

public class Solution {
    public void nextPermutation(int[] A) {
        if(A == null || A.length <= 1) return;
        int i = A.length - 2;
        while(i >= 0 && A[i] >= A[i + 1]) i--; // Find 1st id i that breaks descending order
        if(i >= 0) {                           // If not entirely descending
            int j = A.length - 1;              // Start from the end
            while(A[j] <= A[i]) j--;           // Find rightmost first larger id j
            swap(A, i, j);                     // Switch i and j
        }
        reverse(A, i + 1, A.length - 1);       // Reverse the descending sequence
    }

    public void swap(int[] A, int i, int j) { // 俩交换方法差不多
        // int tmp = A[i];
        // A[i] = A[j];
        // A[j] = tmp;
        A[i] = A[i] + A[j];
        A[j] = A[i] - A[j];
        A[i] = A[i] - A[j];
    }

    public void reverse(int[] A, int i, int j) {
        while(i < j) swap(A, i++, j--);
    }
}
// 题目思路
// 先右边找第一个违反递减规则的数，这个数后面就是递减的了，该数为目标数a
// 再从右边找第一个比目标数a大的数，该数为目标数b
// a b互换，中间的递减序列分别互换，得结果
// 思路相当棒啊~~~~