27 - Next Permutation

最新推荐文章于 2021-01-14 15:08:04 发布

原创最新推荐文章于 2021-01-14 15:08:04 发布 · 406 阅读

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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）。文章详细解释了寻找最尾部顺序对并进行反转的方法。

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

solution: 找出最尾的顺序对, 反转。先从尾往头扫到非递增的数截止（若扫到0，则表示该数最大，立即反转），然后以该数为对比继续从尾往头扫，直到有一个数比他大，交换这两个数，然后立即反转整个从尾递增的数列。

class Solution {
public:
    
    void nextPermutation(vector<int> &num) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if(num.size() == 0)
            return;
        int len = num.size();
        int fir = len -1;
        for( fir; fir >= 1; fir--)
        {
            if( num[fir] > num[fir-1] )
                break;
        }
        
        if( fir > 0 )
        {
            fir --;
            int sec = len -1;
            while( sec >= 0 && num[sec] <= num[fir] )
            {
                sec --;
            }
            //swap fir and sec
            int temp = num[ fir ];
            num[ fir ] = num[ sec ];
            num[ sec ] = temp;
            fir ++;
        }
        
        int end = len - 1;
        while(end > fir)
        {
            int temp = num[ fir ];
            num[ fir ] = num[ end ];
            num[ end ] = temp;
            end--;
            fir++;
        }
        
    }
};