0
点赞
收藏
分享

微信扫一扫

lintcode:3 Sum Closest


Given an array S of n integers, find three integers in S such that the sum is closest to a given number, target. Return the sum of the three integers.

Example
For example, given array S = {-1 2 1 -4}, and target = 1. The sum that is closest to the target is 2. (-1 + 2 + 1 = 2).

Note
You may assume that each input would have exactly one solution.

Challenge
O(n^2) time, O(1) extra space

先排序。再对每个数,在其后面的序列中,设置头尾两个指针,如果三个数的和小于target,头指针向后移动;如果大于target,尾指针向前移。

空间复杂度O(nlogn)+O(n^2)=O(n^2)

其实,要求时间复杂度为O(n^2)说明必须要设置两个指针相向移动,同时这种相向移动要产生作用的前提需要序列是有序的。

class Solution {
public:
/**
* @param numbers: Give an array numbers of n integer
* @param target: An integer
* @return: return the sum of the three integers, the sum closest target.
*/
int threeSumClosest(vector<int> nums, int target) {
// write your code here

sort(nums.begin(),nums.end());

int res = 0;
bool first = true;

for (int i = 0; i<nums.size(); i++){
int pa = i + 1;
int pb = nums.size() - 1;

while (pa<pb){
int sum = nums[i] + nums[pa] + nums[pb];


if (first){
res = sum;
first = false;
}
else if (abs(target - sum)<abs(target - res)){
res = sum;
}

if (res == target){
return res;
}


/*if(sum==target){
return sum;
}else if(abs(target-sum)<abs(target-res)){
res=sum;
}
*/

/* if(res<=target){
pa++;
}else{
pb--;
}
*/
if (sum <= target){
pa++;
}
else{
pb--;
}


}

}

return res;

}


};


举报

相关推荐

0 条评论