18. 4Sum
In this problem, we use four pointers to get no repeating tuples, two for iteration and two for shrinking the solution space.
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;
}
}