leetcode 3Sum

1、
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: 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] ]

题意：给定一个数组，从中找出所有和为零的三个数，要求结果不重复
思路：从头开始遍历，对每一个数找到两个数和为它的相反数，先排序，然后从两头向中间找，注意找出所有情况和去除重复数字，时间复杂度O(n^2)。

class Solution {public:    vector<vector<int>> threeSum(vector<int>& nums) {        vector<vector<int>> res;        std::sort(nums.begin(), nums.end());        for (int i = 0; i < nums.size(); ++i) {            int target = -nums[i];            int front = i + 1;            int back = nums.size() - 1;            while (front < back) {                int sum = nums[front] + nums[back];                if (sum < target)                     front++;                else if (sum > target)                    back--;                else {                    vector<int> tmp(3, 0);                    tmp[0] = nums[i];                    tmp[1] = nums[front];                    tmp[2] = nums[back];                    res.push_back(tmp);                    //跳过重复数字                    while (front < back && nums[front] == tmp[1]) front++;                    while (front < back && nums[back] == tmp[2]) back--;                }            }            //跳过重复数字            while (i + 1 < nums.size() && nums[i + 1] == nums[i]) i++;        }        return res;    }};

2、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).

题意：给定S个整数和一个目标数，在S个整数中找出三个数，使他们的和距离目标数最近，结果唯一。
思路：同上，时间复杂度O(n^2)；

class Solution {public:    int threeSumClosest(vector<int>& nums, int target) {        if (nums.size() < 3) return 0;        int closest = nums[0] + nums[1] + nums[2];        std::sort(nums.begin(), nums.end());        for (int i = 0; i < nums.size() - 2; ++i) {            int front = i + 1, back = nums.size()-1;            while (front < back) {                int tmp = nums[i] + nums[front] + nums[back];                if (tmp == target) return tmp;                if (abs(target - tmp) < abs(target - closest)) closest = tmp;                if (tmp > target) --back;                else ++front;            }            while (i + 1 < nums.size() && nums[i] == nums[i+1]) ++i;        }        return closest;    }};