三个数循环太复杂
确定一个数,搜索另两个
先排序,之后就确定了搜索的策略
if(tp>target)
while (l < r && nums[r] == nums[--r]);
else if (tp<target)
while (l < r && nums[l] == nums[++l]);
else
return target;
因为题目确定恰好有一组最优解,所以不用判断特殊情况
一旦temp=target,直接返回
class Solution {
public:
int threeSumClosest(vector<int>& nums, int target) {
sort(nums.begin(),nums.end());
int ret=nums[0]+nums[1]+nums[2];
int len=nums.size();
int distance =abs(target-ret);
if(ret==target)
return target;
for (int i = 0; i < len - 2; i++) {
if(ret==target)
return target;
if((i>0)&&nums[i]==nums[i-1])
continue;
int l = i + 1;
int r = len - 1;
while (l < r) { //多组
int tp = nums[i] + nums[l] + nums[r];
if (abs(target-tp) < distance){
ret=tp;
distance=abs(target-tp);
}
if(tp>target)
while (l < r && nums[r] == nums[--r]);
else if (tp<target)
while (l < r && nums[l] == nums[++l]);
else
return target;
}
}
return ret;
}
};