Leetcode 题目解析之 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：

• 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)

对数组进行排序；

start从0到n-1，对numstart求另外两个数，这里用mid和end；

mid指向start+1，q指向结尾。sum = numstart + nummid+ numend；

利用加逼定理求解，终止条件是mid == end；

顺带去重

去重：

如果start到了start+1，numstart == numstart - 1，则求出的解肯定重复了；

如果mid++，nummid = nummid - 1，则求出的解肯定重复了。

public List<List<Integer>> threeSum(int[] nums) {
    if (nums == null || nums.length < 3) {
        return new ArrayList<List<Integer>>();
    }
    Set<List<Integer>> set = new HashSet<List<Integer>>();
    Arrays.sort(nums);
    for (int start = 0; start < nums.length; start++) {
        if (start != 0 && nums[start - 1] == nums[start]) {
            continue;
        }
        int mid = start + 1, end = nums.length - 1;
        while (mid < end) {
            int sum = nums[start] + nums[mid] + nums[end];
            if (sum == 0) {
                List<Integer> tmp = new ArrayList<Integer>();
                tmp.add(nums[start]);
                tmp.add(nums[mid]);
                tmp.add(nums[end]);
                set.add(tmp);
                while (++mid < end && nums[mid - 1] == nums[mid])
                    ;
                while (--end > mid && nums[end + 1] == nums[end])
                    ;
            }
            else if (sum < 0) {
                mid++;
            }
            else {
                end--;
            }
        }
    }
    return new ArrayList<List<Integer>>(set);
}