leetcode--16. 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.

具体思路为：

step1：固定一个数字，for循环从0到len-2

step2：另外两个数字采用TwoClosest的方法，利用2 pointer进行求解。

时间复杂度为（O(N^3)）

class Solution {
public:

int twoClosest(vector<int>& nums, int start, int len, int tar) {
int min = INT_MAX;
for(int i = start; i<len; i++){
int other = tar - nums[i];
for(int j = i+1; j<len; j++){
min = abs(min) > abs(nums[j] - other) ? (nums[j] - other) : min;
}
}
return min;
}

int threeSumClosest(vector<int>& nums, int target) {
int len = nums.size();
int min = INT_MAX;
for(int i = 0; i< len; i++){
int sum = target - nums[i];
//two closest by input sum;
int ret = twoClosest(nums, i+1, len, sum);
min = abs(min) > abs(ret) ? ret : min;
}
return target + min;
}
};

利用two pointers可以将时间复杂度降低到n^2:

在Two closest处进行了优化。

class Solution {
public:

int twoClosest(vector<int>& nums, int start, int len, int tar) {
int i = start;
int j = len;
int min = tar - (nums[i] + nums[j]);
while(i<j){
int tmp = nums[i] + nums[j];
int new_min = abs(tar-tmp);
if(abs(min) > new_min) min = tar - tmp;
if(tar == tmp) return 0;
else{
if(tar > tmp) i++;
else j--;
}
}
return min;
}

int threeSumClosest(vector<int>& nums, int target) {
int len = nums.size();
sort(nums.begin(), nums.end());
int min = target - (nums[0] + nums[1] + nums[2]);
for(int i = 0; i< len-2; i++){
int sum = target - nums[i];
//two closest by input sum;
int ret = twoClosest(nums, i+1, len-1, sum);
min = abs(min) > abs(ret) ? ret : min;
}
return target - min;
}
};

答案中更简洁的方法：

class Solution {
public:
int threeSumClosest(vector<int>& nums, int target) {

sort(nums.begin(), nums.end());
int closest = nums[0] + nums[1] + nums[nums.size() - 1];
for(int i = 0; i < nums.size() - 2; i++){
int left = i+1, right = nums.size() - 1;
while(left < right){
int sum = nums[left] + nums[right] + nums[i];
if(abs(sum - target) < abs(closest - target))
closest = sum;
if(sum > target) right--;
else if(sum < target) left++;
else break;
}

}
return closest;
}
};