‘3Sum’ in LeetCode (Medium)
📌 Problem
https://leetcode.com/problems/3sum/
📌 Answer
The point
- Sort array first
- Used Two-pointers technique
- i and j
More detail about ‘Two pointers algorithm’
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class Solution {
public List<List<Integer>> threeSum(int[] nums) {
Arrays.sort(nums);
List<List<Integer>> list = new ArrayList<>();
for (int i = 0; i < nums.length - 2; i++) {
// avoid duplicates
if (i > 0 && (nums[i] == nums[i-1])) continue;
// start : i, j , , , , , k
int j = i + 1;
int k = nums.length-1;
while (j < k) {
int sum = nums[i] + nums[j] + nums[k];
if (sum == 0) {
list.add(Arrays.asList(nums[i], nums[j], nums[k]));
// move again to make new list
j++;
k--;
// avoid duplicates
while ((j < k) && (nums[j] == nums[j - 1])) j++;
while ((j < k) && (nums[k] == nums[k + 1])) k--;
}
else if (sum > 0) k--;
else if (sum < 0) j++;
}
}
return list;
}
}
/**
[Question]
- return all the triplets [nums[i], nums[j], nums[k]]
- condition
- i != j, i != k, j != k (index)
- nums[i] + nums[j] + nums[k] == 0 (value)
- the solution set must not contain duplicate triplets.
**/
The source : https://leetcode.com/problems/3sum/discuss/7631/Simple-Java-Solution-Without-using-HashSet
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