20
04/2015
[LeetCode] 3Sum Closest
3Sum 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. You may assume that each input would have exactly one solution.
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).解题思路：
类似于3Sum。首先给数组排序，用两个变量分别记录当前最小距离和最近的和。代码如下，时间复杂度为O(n^2)
class Solution {
public:
int threeSumClosest(vector<int> &num, int target) {
int len=num.size();
if(len<3){
return 0;
}
std::sort(num.begin(), num.end());
int closest=num[0]+num[1]+num[2];
int minDis = abs(closest-target);
for(int i=0; i<len; i++){
int start=i+1, end=len-1;
while(start<end){
int sum=num[i]+num[start]+num[end];
int dis=abs(sum-target);
if(dis<minDis){
minDis=dis;
closest=sum;
}
if(sum>target){
end--;
}else{
start++;
}
}
}
return closest;
}
int abs(int a){
return a>=0?a:-a;
}
};二次刷题（2015-08-18）
class Solution {
public:
int threeSumClosest(vector<int>& nums, int target) {
int len = nums.size();
if(len < 3){
return 0;
}
std::sort(nums.begin(), nums.end());
int minDis = INT_MAX;
int result = nums[0] + nums[1] + nums[2];
for(int i = 0; i<len; i++){
int start = i + 1, end = len - 1;
while(start < end){
int sum = nums[i] + nums[start] + nums[end];
int dis = abs(sum - target);
if(dis < minDis){
result = sum;
minDis = dis;
}
if(sum < target){
start++;
}else{
end--;
}
}
}
return result;
}
};转载请注明：康瑞部落 » [LeetCode] 3Sum Closest
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