【LeetCode】3Sum,3Sum Closest 题解报告

【题目1】3Sum
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]
]

【解析】
根据题目的意思，需要在一个数组中找到所有的三元组，使这3个数的和为0.跟Two Sum这道题很像。只不过这道题是找出所有可行的结果，并且结果集中不能存在重复。
这道题依然有两种思路，最直接的想法是进行3重循环，遍历所有的可能，这样的时间效率是O(n^3).显然，时间效率过高。
思路2：先对数组进行排序（递增），效率是O(nlogn),然后先确定第一个值，再在有序的序列中查找两个满足要求的数。由于是有序的，所以查找效率是O(n)，因此，总体的时间效率是O(n^2).

public static List<List<Integer>> threeSum(int[] nums) {    Arrays.sort(nums);    Set<List<Integer>> res = new HashSet<List<Integer>>();    for(int i=0;i<nums.length;i++){        int target = 0 - nums[i];        int j = i+1, k=nums.length-1;        while(j<k){            if(nums[k] + nums[j] == target){                List<Integer> tmp = new ArrayList<Integer>();                tmp.add(nums[i]);                tmp.add(nums[j]);                tmp.add(nums[k]);                res.add(tmp);                j++;                k--;            }            else if(nums[k] + nums[j] < target)                j++;            else                k--;        }    }               return new ArrayList<List<Integer>>(res);}

【题目2】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).

【解析】
这道题跟上面的类似，解题思路是一样的。先对数组进行排序，然后确定第一个数，再在有序的数组中遍历找到最接近目标值的数。

public static int threeSumClosest(int[] nums, int target) {      if(nums == null || nums.length < 3)          return 0;       Arrays.sort(nums);       int res = 0;       int diff = Integer.MAX_VALUE;       int len = nums.length;       for(int i=0;i<len-2;i++){            int j = i+1, k=len-1;            while(j<k){                int sum = nums[i] + nums[j] + nums[k];                if(Math.abs(sum - target) < diff){                    diff = Math.abs(sum - target);                    res = sum;                }                if(sum < target){                    j++;                }else if(sum > target){                    k--;                }else{                    return sum;                }            }                     }       return res;   }