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class Solution {
public:
void nextPermutation(vector<int>& nums) {
int i = nums.size() - 2, j = nums.size() - 1;
while(nums[i] >= nums[i+1]) i--;
if(i >= 0){
while(nums[j] <= nums[i]) j--;//这里一定要是小于等于，因为等于也不可以交换，否则就不是下一个排列组合了
swap(nums[j], nums[i]);
}
reverse(nums.begin() + i + 1, nums.end());
}
};
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31. Next Permutation 题目:Implement next permutation, which...
题目： 31. Next Permutation Implement next permutation, whic...
LeetCode中与Permutations相关的共有四题： 31. Next Permutation 46....
Description Implement next permutation, which rearranges ...
Description: Implement next permutation, which rearranges...
题目 Implement next permutation, which rearranges numbers i...
My Submissions Total Accepted: 91019Total Submissions: 32...
题目 分析 题目要求是给出一个数字排列中按字典序排列的下一个更大的排列。如果没有更大的则给出最小的，比如1,2,3...
本文标题：31. Next Permutation
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