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18 4Sum
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42 Trapping Rain Water
76 Minimum Window Substring
80 Remove Duplicates from Sorted Array II
88 Merge Sorted Array
125 Valid Palindrome
159 Longest Substring with At Most Two Distinct Characters
167 Two Sum II - Input array is sorted
202 Happy Number
209 Minimum Size Subarray Sum
259 3Sum Smaller
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340 Longest Substring with At Most K Distinct Characters
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349 Intersection of Two Arrays
350 Intersection of Two Arrays II
360 Sort Transformed Array
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18 4Sum

18. 4Sum​

1. Question

Given an arraySofnintegers, are there elementsa,b,c, anddinSsuch thata+b+c+d= target? Find all unique quadruplets in the array which gives the sum of target.
Note:The solution set must not contain duplicate quadruplets.
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For example, given array S = [1, 0, -1, 0, -2, 2], and target = 0.
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​
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A solution set is:
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[
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[-1, 0, 0, 1],
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[-2, -1, 1, 2],
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[-2, 0, 0, 2]
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]
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2. Implementation

(1) Two Pointers
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class Solution {
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public List<List<Integer>> fourSum(int[] nums, int target) {
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List<List<Integer>> res = new ArrayList<>();
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​
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Arrays.sort(nums);
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​
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for (int i = 0; i < nums.length - 3; i++) {
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if (i > 0 && nums[i - 1] == nums[i]) continue;
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​
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for (int j = i + 1; j < nums.length - 2; j++) {
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if (j > i + 1 && nums[j - 1] == nums[j]) continue;
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​
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int k = j + 1, m = nums.length - 1;
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​
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while (k < m) {
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int sum = nums[i] + nums[j] + nums[k] + nums[m];
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​
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if (sum == target) {
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res.add(Arrays.asList(nums[i], nums[j], nums[k], nums[m]));
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++k;
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--m;
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​
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while (k < m && nums[k - 1] == nums[k]) ++k;
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while (k < m && nums[m] == nums[m + 1]) --m;
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}
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else if (sum < target) {
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++k;
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}
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else {
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--m;
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}
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}
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}
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}
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return res;
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}
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}
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3. Time & Space Complexity

Two Pointers: 时间复杂度O(n^3)
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Contents
18. 4Sum
1. Question
2. Implementation
3. Time & Space Complexity