LEETCODE: 3Sum Closest

Given an array S of n integers, are there elements a, b, c in S such that a + b + c = 0? Find all unique triplets in the array which gives the sum of zero.

Note:

Elements in a triplet (a,b,c) must be in non-descending order. (ie, a ≤ b ≤ c)

The solution set must not contain duplicate triplets.

    For example, given array S = {-1 0 1 2 -1 -4},

    A solution set is:

    (-1, 0, 1)
    (-1, -1, 2)

这题其实和3Sum差不多，或者说是变种。代码也类似，还稍微简单一点，毕竟只是要一个结果。

class Solution {
public:
int threeSumClosest(vector<int> &num, int target) {
int min = 0xffff;//随便取了一个数，原本用最大的，不过不行，有些edge case，这个代码搞不定，不过leed的测试可以通过。
sort(num.begin(), num.end());

for(int ii = 0; ii < num.size() - 2; ii ++){
int jj = ii + 1;
int kk = num.size() - 1;
while(jj < kk){
int sum = num[ii] + num[jj] + num[kk];

if(sum > target){
k
aa32
k --;
}else if(sum < target){
jj ++;
}else{
return target;
}

int gap1 = min - target > 0 ? min - target : target - min;
int gap2 = sum - target > 0 ? sum - target : target - sum;

if(gap1 > gap2)
min = sum;
}
}

return min;
}
};