Leetcode-15. 3Sum
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.
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.
<解題>
class Solution {
public List<List<Integer>> threeSum(int[] nums) {
List<List<Integer>> ans = new ArrayList<>();
Arrays.sort(nums); // nums = [-4,-1,-1,0,1,2]
for (int i = 0; i < nums.length - 2; i++) {
// Skip duplicate elements for i
if (i > 0 && nums[i] == nums[i - 1]) {
continue;
}
int j = i + 1;
int k = nums.length - 1;
while (j < k) {
int sum = nums[i] + nums[j] + nums[k];
if (sum == 0) {
// Found a triplet with zero sum
ans.add(Arrays.asList(nums[i], nums[j], nums[k]));
// Skip duplicate elements for j
while (j < k && nums[j] == nums[j + 1]) {
j++;
}
// Skip duplicate elements for k
while (j < k && nums[k] == nums[k - 1]) {
k--;
}
// Move the pointers
j++;
k--;
} else if (sum < 0) {
// Sum is less than zero, increment j to increase the sum
j++;
} else {
// Sum is greater than zero, decrement k to decrease the sum
k--;
}
}
}
return ans;
}
}
Time: O(n^2)
- 數組排序O(n log n)
- 外層O(n)、內層：雙指針O(n)->)(n^2)
- -> 即 O(n log n) + O(n) * O(n) = O(n^2) Space: O(n^2)