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java_three_sum

Given an integer array nums, return all the triplets [nums[i], nums[j], nums[k]] such that i != j, i != k, and j != k, and nums[i] + nums[j] + nums[k] == 0.

Notice that the solution set must not contain duplicate triplets.

Examples

Example 1:

Input: nums = [-1,0,1,2,-1,-4]
Output: [[-1,-1,2],[-1,0,1]]
Explanation:
nums[0] + nums[1] + nums[2] = (-1) + 0 + 1 = 0.
nums[1] + nums[2] + nums[4] = 0 + 1 + (-1) = 0.
nums[0] + nums[3] + nums[4] = (-1) + 2 + (-1) = 0.
The distinct triplets are [-1,0,1] and [-1,-1,2].
Notice that the order of the output and the order of the triplets does not matter.

Example 2:

Input: nums = [0,1,1]
Output: []
Explanation: The only possible triplet does not sum up to 0.

Example 3:

Input: nums = [0,0,0]
Output: [[0,0,0]]
Explanation: The only possible triplet sums up to 0.

Constraints:

3 <= nums.length <= 3000
$-10^5$ <= nums[i] <= $10^5$

解析

給定一個整數陣列 nums

要求寫一個演算法找在所有在 nums 中3個和= 0 數字的所有不重複複組合

已知目標是找到 i, j , k , i < j < k , 使得 nums[i] + nums[j] + nums[k] =0

可以注意到當 nums[j] 確定時, nums[i] + nums[k] 只能 - nums[j]

所以可透過類似 two sum 的作法

另外為了可以使用 two pointer 來縮小範圍

所以將原本的nums 做 sort 來做由小到大的排列

初始化 start = 0, index = 1, end = len(nums) - 1, result = []

當針對所有可能的 index 做檢查(使用 two pointer)

為了避免重複檢查

所以如果 nums[index] = nums[index-1] 時且 index > 1 代表已經nums[index]找過了

這時需要把 start = index - 1 讓搜索範圍變小

而對於 start < index 且 end > index 時

為了避免重複檢查

當nums[start] = nums[start-1] 且 start > 0 代表此搜索範圍已經找過了

這時需要把 start += 1 重新搜索

當nums[end] = nums[end+1] 且 end < len(nums) - 1 代表此搜索範圍已經找過了

這時需要把 end -= 1 重新搜索

令 addNum = nums[start] + nums[index] + nums[end]

當 addNum = 0 代表找到符合條件的組合

把 [nums[start], nums[index], nums[end]] 加入 result , 更新 start += 1, end -= 1

當 addNum > 0 因為 start 已經是最小值所以只能把 end 往左移

更新 end -= 1

當 addNum < 0 因為 end 已經是最大值所以只能把 start 往右移

更新 start += 1

程式碼

import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;

public class Solution {
  public List<List<Integer>> threeSum(int[] nums) {
    int n = nums.length;
    if (n < 3) {
      return null;
    }
    Arrays.sort(nums);
    List<List<Integer>> result = new ArrayList<>();
    for (int pivot = 1; pivot < n - 1; pivot++) {
      int start = 0, end = n - 1;
      if (pivot > 1 && nums[pivot] == nums[pivot-1]) {
        start = pivot - 1;
      }
      while( start < pivot && pivot < end) {
        if (start > 0 && nums[start] == nums[start-1]) {
          start++;
          continue;
        }
        if (end < n-1 && nums[end] == nums[end+1]) {
          end--;
          continue;
        }
        int sum = nums[start] + nums[pivot] + nums[end];
        if (sum == 0) {
          result.add(List.of(nums[start], nums[pivot], nums[end]));
          start++;
          end--;
        } else if (sum > 0) {
          end--;
        } else {
          start++;
        }
      }
    }
    return result;
  }
}