• 15. 3Sum


    Given an array nums of n integers, are there elements abc in nums 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.

    Example:

    Given array nums = [-1, 0, 1, 2, -1, -4],
    
    A solution set is:
    [
      [-1, 0, 1],
      [-1, -1, 2]
    ]

    sort the input array and then run through all indices of a possible first element of a triplet.

    for each possible first element, use two pointers like in 2sum problem

    skip equal elements to avoid duplicates in the answer without a set

    注意:在skip same element的时候,l和r的方法是相反的,应该判断nums[l] = nums[l-1]和nums[r] = nums[r+1]

    时间:O(N^2),空间:O(1)

    class Solution {
        public List<List<Integer>> threeSum(int[] nums) {
            Arrays.sort(nums);
            List<List<Integer>> res = new ArrayList<>();
            for(int i = 0; i + 2 < nums.length; i++) {
                if(i == 0 || (i > 0 && nums[i] != nums[i - 1])) {
                    int l = i + 1, r = nums.length - 1, target = -nums[i];
                    while(l < r) {
                        if(nums[l] + nums[r] == target) {
                            res.add(Arrays.asList(nums[i], nums[l], nums[r]));
                            l++;r--;
                            while(l < r && nums[l] == nums[l - 1])
                                l++;
                            while(l < r && nums[r] == nums[r + 1])
                                r--;
                        } else if(nums[l] + nums[r] > target) {
                            r--;
                        } else {
                            l++;
                        }
                    }
                }
            }
            return res;
        }
    }
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  • 原文地址:https://www.cnblogs.com/fatttcat/p/10039480.html
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