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【Leetcode】Next Permutation

2024-07-18 13:22:50   214人阅读

 

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

 

1st（1 tries）
class Solution {public:    void nextPermutation(vector<int> &num) {        // Note: The Solution object is instantiated only once and is reused by each test case.        next_permutation(num.begin(),num.end());    }};

　　

2nd（ 7 tries）
class Solution {public:    void nextPermutation(vector<int> &num) {        vector<int>::iterator last = num.end() - 1;        if(last == num.begin())            return;        vector<int>::iterator prelast = last-1;        while(last != num.begin() && *prelast >= *last) {            prelast--;            last--;        }        if(last == num.begin()) {            sort(num.begin(),num.end());            return;        }        //from last to end() and get the minest        vector<int>::iterator minest = last;        for(vector<int>::iterator iter = last + 1;iter != num.end();++iter) {            if(*iter > *prelast && *iter <= *minest) {                minest = iter;            }        }        swap(*minest,*prelast);        reverse(last,num.end());    }};

　　


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