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leetcode 15. 3sum

admin 11月 12, 2020 0

15. 3Sum

Difficulty: Medium

Given an array nums of n integers, are there elements a, b, c 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:

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Given array nums = [-1, 0, 1, 2, -1, -4],

A solution set is:
[
  [-1, 0, 1],
  [-1, -1, 2]
]

Solution

Language: Java

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class  {
      public List<List<Integer>> threeSum(int[] nums) {
            List<List<Integer>> result = new ArrayList<>();
            if (nums == null || nums.length == 0) {
                  return result;
           }
            Arrays.sort(nums);
            for (int i = 0; i < nums.length && nums[i] <= 0; i++) {
                  for (int j = i + 1; j < nums.length; j++) {
                        int curSum = nums[i] + nums[j];
                        int index = Arrays.binarySearch(nums, j + 1, nums.length, -curSum);
                        if (index >= 0) {
                              List<Integer> tmp = new ArrayList<>();
                              tmp.add(nums[i]);
                              tmp.add(nums[j]);
                              tmp.add(nums[index]);
                              result.add(tmp);
                       }
                        while (j + 1 < nums.length && nums[j+1] == nums[j]) {
                              j++;
                       }
                 }
                  while (i + 1 < nums.length && nums[i+1] == nums[i]) {
                        i++;
                 }
           }
            return result;
     }
}

两根指针

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class  {
    public List<List<Integer>> threeSum(int[] nums) {
        List<List<Integer>> result = new ArrayList<>();
        if (nums == null || nums.length == 0) {
            return result;
        }
        Arrays.sort(nums);
        int n = nums.length;
        int target = 0;
        for (int i = 0; i < n - 2; i++) {
            if (i > 0 && nums[i] == nums[i - 1]) {
                continue;
            }
            int j = i + 1;
            int k = n - 1;
            while (j < k) {
                int res = nums[i] + nums[j] + nums[k];
                if (res == 0) {
                    result.add(Arrays.asList(nums[i], nums[j++], nums[k--]));
                    while (j < k && nums[j] == nums[j - 1]) {
                        j++;
                    }
                    while (j < k && nums[k] == nums[k + 1]) {
                        k--;
                    }
                } else if (res > 0) {
                    k--;
                } else {
                    j++;
                }
            }
        }
        return result;
    }
}
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