Given an array S of n integers, are there elements a, b, c, and d in S such that a + b + c + d = target? Find all unique quadruplets in the array which gives the sum of target.
For example, given array S = {1 0 -1 0 -2 2}, and target = 0.
A solution set is:
(-1, 0, 0, 1)
(-2, -1, 1, 2)
(-2, 0, 0, 2)
找出一個序列中所有和為target的四元組庸蔼。
循環(huán)的3Sum筹燕。
public class Solution {
public List<List<Integer>> fourSum(int[] nums, int target) {
List result = new ArrayList<List<Integer>>();
if (nums == null || nums.length < 4) return result;
Arrays.sort(nums);
for (int i = 0; i < nums.length - 3; i++) {
if (i > 0 && nums[i] == nums[i-1]) continue;
for (int j = i + 1; j < nums.length - 2; j++) {
if (j > i + 1 && nums[j] == nums[j-1]) continue;
int left = j + 1;
int right = nums.length - 1;
while (left < right) {
int tmp = nums[i] + nums[j] + nums[left] + nums[right];
if (tmp == target) {
List l = new ArrayList<Integer>();
l.add(nums[i]);
l.add(nums[j]);
l.add(nums[left]);
l.add(nums[right]);
result.add(l);
while (left < right && nums[++left] == nums[left-1]);
while (left < right && nums[--right] == nums[right+1]);
} else if (tmp > target) while (left < right && nums[--right] == nums[right+1]);
else while (left < right && nums[++left] == nums[left-1]);
}
}
}
return result;
}
}