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;    }};