15. 3Sum
In this problem, we use three pointers to get no repeating tuples, one for iteration and two for shrinking the solution space.
class Solution {
public List<List<Integer>> threeSum(int[] nums) {
if (nums == null || nums.length < 3) {
return new ArrayList<>();
}
// quick sort
Arrays.sort(nums);
List<List<Integer>> lists = new ArrayList<>();
for (int i = 0; i < nums.length; i++) {
// no 3sum tuple in the rest of the sorted array
if (nums[i] > 0) {
return lists;
}
// If nums[i] is duplicated with nums[i - i], skips this round of loop.
if (i > 0 && nums[i] == nums[i - 1]) {
continue;
}
int left = i + 1;
int right = nums.length - 1;
while (left < right) {
if (nums[i] + nums[left] + nums[right] == 0) {
lists.add(Arrays.asList(nums[i], nums[left], nums[right]));
// skips duplicated elements (if it exists)
while (left < right && nums[left] == nums[left + 1]) {
left = left + 1;
}
while (left < right && nums[right] == nums[right - 1]) {
right = right - 1;
}
// moves to the next element
left = left + 1;
right = right - 1;
} else if (nums[i] + nums[left] + nums[right] > 0) {
right = right - 1;
} else {
left = left + 1;
}
}
}
return lists;
}
}
16. 3Sum Closest
class Solution {
public int threeSumClosest(int[] nums, int target) {
// corner cases
if (nums.length == 1) {
return nums[0];
}
if (nums.length == 2) {
int diff1 = Math.abs(nums[0] - target);
int diff2 = Math.abs(nums[1] - target);
return Math.min(diff1, diff2);
}
Arrays.sort(nums);
int closestSum = nums[0] + nums[1] + nums[nums.length - 1];
for (int i = 0; i < nums.length; i++) {
// If nums[i] is duplicated with nums[i - i], skip this round of loop.
if (i - 1 >= 0 && nums[i] == nums[i - 1]) {
continue;
}
int left = i + 1;
int right = nums.length - 1;
while (left < right) {
int sum = nums[i] + nums[left] + nums[right];
if (sum > target) {
right--;
} else if (sum < target){
left++;
} else {
return sum;
}
int diffOld = Math.abs(closestSum - target);
int diffNew = Math.abs(sum - target);
if (diffNew < diffOld) {
closestSum = sum;
}
}
}
return closestSum;
}
}
259. 3Sum Smaller
In this problem, we should count the numbers of tuples which is smaller than the target, and we would not skip those duplicated tuples.
class Solution {
public int threeSumSmaller(int[] nums, int target) {
if (nums.length <= 2) {
return 0;
}
Arrays.sort(nums);
int count = 0;
for (int i = 0; i < nums.length; i++) {
int left = i + 1;
int right = nums.length - 1;
while (left < right) {
int sum = nums[i] + nums[left] + nums[right];
if (sum < target) {
count += right - left;
left++;
} else {
right--;
}
}
}
return count;
}
}
18. 4Sum
In this problem, we use four pointers to get no repeating tuples, two for iteration and two for shrinking the solution space.
import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
class Solution {
public List<List<Integer>> fourSum(int[] nums, int target) {
// corner case
List<List<Integer>> result = new ArrayList<>();
if (nums == null || nums.length < 4) {
return result;
}
Arrays.sort(nums);
int length = nums.length;
// the range of i is [0, length - 3),
// that is because there are always three elements
// nums[j], nums[left] and nums[right] lining up on the right of nums[i],
// thus the right excluding endpoint is (length - 3)
for (int i = 0; i < length - 3; i++) {
// skips duplicated i
if (i > 0 && nums[i] == nums[i - 1]) {
continue;
}
// if the minimum of i loops is larger than target, skips the rest
int iMin = nums[i] + nums[i + 1] + nums[i + 2] + nums[i + 3];
if (iMin > target) {
break;
}
// if the maximum of i loops is smaller than target, ignore this round
int iMax = nums[i] + nums[length - 3] + nums[length - 2] + nums[length - 1];
if (iMax < target) {
continue;
}
// the range of j is [i + 1, length - 2),
// that is because there are always two elements
// nums[left] and nums[right] lining up on the right of nums[j],
// thus the right excluding endpoint is (length - 2)
for (int j = i + 1; j < length - 2; j++) {
if (j > i + 1 && nums[j] == nums[j - 1]) {
continue;
}
int left = j + 1;
int right = length - 1;
// for a fixed nums[i],
// if the minimum of j loops is larger than target, ignore this round
int jMin = nums[i] + nums[j] + nums[left] + nums[left + 1];
if (jMin > target) {
continue;
}
// for a fixed nums[i],
// if the maximum of j loops is smaller than target, ignore this round
int jMax = nums[i] + nums[j] + nums[right - 1] + nums[right];
if (jMax < target) {
continue;
}
while (left < right) {
int sum = nums[i] + nums[j] + nums[left] + nums[right];
if (sum == target) {
result.add(
Arrays.asList(nums[i], nums[j], nums[left], nums[right]));
while (left < right && nums[left] == nums[left - 1]) {
left++;
}
while (left < right && nums[right] == nums[right + 1]) {
right--;
}
left++;
right--;
} else if (sum > target) {
right--;
} else {
left++;
}
}
}
}
return result;
}
}
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