LeetCode 18. 4Sum [Medium]

题目来源:https://leetcode.com/problems/4sum/

题目难度:Medium

解答1[Java]

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class Solution {
public List<List<Integer>> fourSum(int[] num, int target) {
ArrayList<List<Integer>> ans = new ArrayList<>();
if (num.length < 4)
return ans;
Arrays.sort(num);
for (int i = 0; i < num.length - 3; i++) {
if (num[i] + num[i + 1] + num[i + 2] + num[i + 3] > target)
break; // first candidate too large, search finished
if (num[i] + num[num.length - 1] + num[num.length - 2] + num[num.length - 3] < target)
continue; // first candidate too small
if (i > 0 && num[i] == num[i - 1])
continue; // prevents duplicate result in ans list
// already sorted, only need to check the adjacent element

for (int j = i + 1; j < num.length - 2; j++) {
if (num[i] + num[j] + num[j + 1] + num[j + 2] > target)
break; // second candidate too large
if (num[i] + num[j] + num[num.length - 1] + num[num.length - 2] < target)
continue; // second candidate too small
if (j > i + 1 && num[j] == num[j - 1])
continue; // prevents duplicate results in ans list
int low = j + 1, high = num.length - 1;
while (low < high) {
int sum = num[i] + num[j] + num[low] + num[high];
if (sum == target) {
ans.add(Arrays.asList(num[i], num[j], num[low], num[high]));
while (low < high && num[low] == num[low + 1])
low++; // skipping over duplicate on low
while (low < high && num[high] == num[high - 1])
high--; // skipping over duplicate on high
low++;
high--;
}
// move window
else if (sum < target)
low++;
else
high--;
}
}
}
return ans;
}
}