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LeetCode :: Next Permutation

2024-07-16 09:40:00   225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

class Solution {
public:
    void reverse(vector<int>&num, int left, int right){
        for (int i = left, j = right; i < j; i++, j--){
            swap (num[i], num[j]);
        }  
    }
    
    void nextPermutation(vector<int> &num) {
        int n = num.size();
        if (n == 0 || n == 1) return;
        int i, ii, j;
        
        for (i = n - 1; i > 0; i--){
            if (num[i] > num[i - 1])
                break;
        }
        ii = i;          //已经包含了是最减低到最小的情况；
        --i;
        
        for (j = n - 1; j >= ii; j--){
            if (num[j] > num[i]){    //找出大于i位置的最小值，利用已经是一个降序的性质；
                swap(num[j], num[i]);
                break;
            }
        }
        reverse(num, ii, n - 1);    
    }
};


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