注意:答案中不可以包含重复的三元组。
leetcode链接
示例 1: 输入:nums = [-1,0,1,2,-1,-4] 输出:[[-1,-1,2],[-1,0,1]]
示例 2: 输入:nums = [] 输出:[]
示例 3: 输入:nums = [0] 输出:[]
思路:先给数组进行从小到大排序。然后定义三个指针:first,second,third,分别代表三元组的三个下标。
first下标从0开始,到nums.size()-1结束。
second下标从first+1开始,third下标从nums.size()-1开始,second要始终小于third。
如果nums[first]+nums[second]+nums[third]>0,说明三数之和大于0,此时third应该往左走,使三数之和减小。
如果nums[first]+nums[second]+nums[third]<0,说明三数之和小于0,此时second应该往右走,使三数之和增加。直到second和third相遇。
另外,要注意去重。
代码:
class Solution { //三数之和
public:
vector> threeSum(vector& nums) {
vector> result;
sort(nums.begin(), nums.end());
if (nums.size() < 3)
return result;
for (int first = 0; first < nums.size(); first++) {
if (nums[first] > 0)
return result;
if (first > 0 && nums[first] == nums[first - 1])
continue;
int second = first + 1;
int third = nums.size() - 1;
while (second < third){
if (nums[first] + nums[second] + nums[third] > 0) {
third--;
while (second < third && nums[third] == nums[third + 1])
third--;
}
else if (nums[first] + nums[second] + nums[third] < 0) {
second++;
while (second < third && nums[second] == nums[second - 1])
second++;
}
else {
result.push_back(vector{nums[first], nums[second], nums[third]});
while (second < third && nums[third] == nums[third - 1])
third--;
while (second < third && nums[second] == nums[second + 1])
second++;
second++;
third--;
}
}
}
return result;
}
};



