LeetCode 31. Next Permutation

最新推荐文章于 2020-10-03 00:35:45 发布

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本文介绍了一个Java实现的下一个排列算法，该算法用于将数组中的整数重新排列成字典序中下一个更大的排列。如果没有这样的排列，则将其变为最小的排列（即升序排列）。文章详细展示了如何找到第一个打破降序排列的元素，并通过交换和反转操作得到结果。

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

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) {
        int tmp = A[i];
        A[i] = A[j];
        A[j] = tmp;
}

    public void reverse(int[] A, int i, int j) {
        while(i < j) swap(A, i++, j--);
    }
}