【LeetCode】4Sum 解题报告
【题目】
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.
Note:
- Elements in a quadruplet (a,b,c,d) must be in non-descending order. (ie, a ≤ b ≤ c ≤ d)
- The solution set must not contain duplicate quadruplets.
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)
【解析】
3Sum和 3Sum Closest 的扩展,相同思路,加强理解。
K Sum 问题的时间复杂度好像为 O(n^(k-1)) ?!假设有更好的,欢迎不吝赐教!
【Java代码】
public class Solution {
List<List<Integer>> ret = new ArrayList<List<Integer>>(); public List<List<Integer>> fourSum(int[] num, int target) {
if (num == null || num.length < 4) return ret;
Arrays.sort(num);
int len = num.length;
for (int i = 0; i < len-3; i++) {
if (i > 0 && num[i] == num[i-1]) continue;
for (int j = i+1; j < len-2; j++) {
if (j > i+1 && num[j] == num[j-1]) continue;
findTwo(num, j+1, len-1, target, num[i], num[j]);
}
}
return ret;
} public void findTwo(int[] num, int begin, int end, int target, int a, int b) {
if (begin < 0 || end >= num.length) return;
int l = begin, r = end;
while (l < r) {
if (a+b+num[l]+num[r] < target) {
l++;
} else if (a+b+num[l]+num[r] > target) {
r--;
} else {
List<Integer> ans = new ArrayList<Integer>();
ans.add(a);
ans.add(b);
ans.add(num[l]);
ans.add(num[r]);
ret.add(ans);
l++;
r--;
while (l < r && num[l] == num[l-1]) l++;
while (l < r && num[r] == num[r+1]) r--;
}
}
}
}
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