Next Permutation
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题目:
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
思路:
举例说明: 对于 1 3 2 6 5 4 ,
变换是观察到下一个应该是1 3 4 2 5 6;
所以其过程是先找出第一个数值,特征是比右边数值小;
再往右寻找最后一个比其小的,不用担心会超过问题,因为后边数是降序排列。
代码:
class Solution {public: void nextPermutation(vector<int>& nums) { int j=nums.size()-2; while(j>=0 && nums[j]>=nums[j+1]){ j--; } if(j>=0){ int i=j+1;//i代表第一个比后面小的数字 的 索引号 while(i<nums.size() && nums[i]>nums[j]){ i++; } i--; swap(nums[i],nums[j]); } reverse(nums,j+1); } void reverse(vector<int>&nums,int index){ int start=index,end=nums.size()-1; while(start<=end){ swap(nums[start],nums[end]); start++;end--; } }};
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