leetcode链接：
https://leetcode.cn/problems/4sum/description/
方案一
思路： 排序 + 对撞指针
时间复杂度：n² log(n)
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| List<List<Integer>> result = new ArrayList<>();
int right,left;
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;
left = j + 1;
right = nums.length - 1;
while (left < right) {
if (nums[i] + nums[j] + nums[left] + nums[right] < 0 && nums[i] > 0) {
break;
}
if (nums[i] + nums[j] + nums[left] + nums[right] > target) {
right--;
} else if (nums[i] + nums[j] + nums[left] + nums[right] == target) {
List<Integer> tem = new ArrayList<Integer>();
tem.add(nums[i]);
tem.add(nums[j]);
tem.add(nums[left]);
tem.add(nums[right]);
result.add(tem);
while (nums[right] == nums[--right]) {
if (right < left + 2)
break;
}
while (nums[left] == nums[++left]) {
if (left > right - 2)
break;
}
} else {
left++;
}
}
}
}
return result;
|
结果：
- 292/292 cases passed (13 ms)
- Your runtime beats 66.36 % of java submissions
- Your memory usage beats 46.39 % of java submissions (41.9 MB)
几乎最优方案
https://leetcode.cn/problems/4sum/solution/si-shu-zhi-he-by-leetcode-solution/
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| class Solution {
public List<List<Integer>> fourSum(int[] nums, int target) {
List<List<Integer>> quadruplets = new ArrayList<List<Integer>>();
if (nums == null || nums.length < 4) {
return quadruplets;
}
Arrays.sort(nums);
int length = nums.length;
for (int i = 0; i < length - 3; i++) {
if (i > 0 && nums[i] == nums[i - 1]) {
continue;
}
if ((long) nums[i] + nums[i + 1] + nums[i + 2] + nums[i + 3] > target) {
break;
}
if ((long) nums[i] + nums[length - 3] + nums[length - 2] + nums[length - 1] < target) {
continue;
}
for (int j = i + 1; j < length - 2; j++) {
if (j > i + 1 && nums[j] == nums[j - 1]) {
continue;
}
if ((long) nums[i] + nums[j] + nums[j + 1] + nums[j + 2] > target) {
break;
}
if ((long) nums[i] + nums[j] + nums[length - 2] + nums[length - 1] < target) {
continue;
}
int left = j + 1, right = length - 1;
while (left < right) {
long sum = (long) nums[i] + nums[j] + nums[left] + nums[right];
if (sum == target) {
quadruplets.add(Arrays.asList(nums[i], nums[j], nums[left], nums[right]));
while (left < right && nums[left] == nums[left + 1]) {
left++;
}
left++;
while (left < right && nums[right] == nums[right - 1]) {
right--;
}
right--;
} else if (sum < target) {
left++;
} else {
right--;
}
}
}
}
return quadruplets;
}
}
作者：LeetCode-Solution
链接：https:
来源：力扣（LeetCode）
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