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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.

本题相当于全排的一种变种。根据当前的排序来找下一个排序结果。先上代码

class Solution {public:    void nextPermutation(vector
   
    & nums) {        int pos = -1;        int len = nums.size();        for(int i = len-1; i>0; i--){            if(nums[i] > nums[i-1]){                pos = i-1;                break;            }        }        if(pos < 0){            reverse(nums, 0, len-1);            return;        }        for(int i = len - 1; i > pos; i--){            if(nums[i] > nums[pos]){                int tmp = nums[pos];                nums[pos] = nums[i];                nums[i] = tmp;                break;            }        }        reverse(nums, pos+1, len-1);    }    void reverse(vector
    
     & num, int begin, int end){        int l = begin;        int r = end;        while(l
    
   

步骤大致分为三步：

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