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| class Solution {
public List<List<Integer>> fourSum(int[] nums, int target) {
if (nums.length < 4) {
return new ArrayList<>();
}
Arrays.sort(nums);
List<List<Integer>> ans = new ArrayList<>();
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, right = nums.length - 1;
long diff = (long) target - nums[i] - nums[j];
while (left < right) {
int currSum = nums[left] + nums[right];
if (currSum < diff) {
left += 1;
} else if (currSum > diff) {
right -= 1;
} else {
ans.add(Arrays.asList(nums[i], nums[j], nums[left], nums[right]));
while (left < right && nums[left] == nums[left + 1])
left += 1;
while (left < right && nums[right] == nums[right - 1])
right -= 1;
left += 1;
right -= 1;
}
}
}
}
return ans;
}
}
|
运用2sum的方法去求4sum. 需要注意的就是如何去重. 挑选nums[i]和nums[j]的时候要避免和之前挑选过的重复. 挑选left和right也是一样.
时间复杂度: O(n^3)
空间复杂度: O(n)
