Next Permutation

最新推荐文章于 2021-03-10 11:29:57 发布

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本文介绍了一个算法，用于在原地将一组数字重新排列成字典序中下一个更大的排列。如果这样的排列不存在，则将其重新排列为最小可能的顺序（即升序）。文章通过具体的代码示例解释了如何实现这一功能。

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

class Solution {
public:
    void nextPermutation(vector<int> &num) {
        int n = num.size();
        int i, j, tmp;
        
        if(n < 2) return;
        
        for(i = n - 2; i >= 0; i--) {
            if(num[i] < num[i + 1]) break;
        }
        for(j = n - 1; j > i; j--) {
            if(num[j] > num[i]) break;
        }
        
        if(i >= 0) {
            tmp = num[i];
            num[i] = num[j];
            num[j] = tmp;
            reverse(num.begin() + i + 1, num.end());
        }
        else reverse(num.begin(), num.end());
    }
};