-
Notifications
You must be signed in to change notification settings - Fork 1
/
LeetCode-18-4Sum.java
110 lines (87 loc) · 4.19 KB
/
LeetCode-18-4Sum.java
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
/*
LeetCode: https://leetcode.com/problems/4sum/
LintCode: http://www.lintcode.com/problem/4sum/
JiuZhang: http://www.jiuzhang.com/solutions/4sum/
ProgramCreek: http://www.programcreek.com/2013/02/leetcode-4sum-java/
Analysis: (Two Pointers solution. The same solution with 3Sum)
1.Sort array
2.P1 scan from 0 to Len-3
2.P2 scan from p1+1 to Len-2
3.P3, P4 scan from P2+1, Len-1, respectively.
方法二:
这道题还有另一个方法,就是先可以用一个hash表记录数组当中任意两个数的和sum = a[x]+a[y],并且记录是哪两个数的和, 这样用o(n^2)的时间复杂度就可以记录任意两个数的和,然后在用两个指针i,j(i<j)遍历数组, 然后在hash表中寻找v-a[i]-a[j]是否存在,并且保证a[i],a[j],a[x],a[y] 不能取相同的数,然后通过i,j两层for循环就可以找到4sum
*/
public class Solution {
// 1.Two Pointers, just like 3Sum
// public List<List<Integer>> fourSum(int[] nums, int target) {
// List<List<Integer>> result = new ArrayList<List<Integer>>();
// if(nums == null && nums.length < 4) return result;
// java.util.Arrays.sort(nums);
// for(int i = 0; i < nums.length - 3; i++){
// if(i != 0 && nums[i - 1] == nums[i]) continue; //skip the duplicates
// for(int j = i + 1; j < nums.length - 2; j++){
// if(j != i + 1 && nums[j - 1] == nums[j]) continue; //skip the duplicates
// int lo = j + 1;
// int hi = nums.length - 1;
// while(lo < hi){
// int sum = nums[i] + nums[j] + nums[lo] + nums[hi];
// if(sum == target){
// List<Integer> temp = new ArrayList<Integer>();
// temp.add(nums[i]);
// temp.add(nums[j]);
// temp.add(nums[lo]);
// temp.add(nums[hi]);
// result.add(temp);
// lo++;
// hi--;
// //skip duplicates
// while(lo < hi && nums[lo - 1] == nums[lo]) lo++;
// while(lo < hi && nums[hi] == nums[hi + 1]) hi--;
// }else if(sum < target){
// lo++;
// }else{
// hi--;
// }
// }
// }
// }
// return result;
// }
// 2.HashTable (How to record the two numbers in HashTable? If key is sum, what can we do if key duplicates, 1+5==2+4)
// public List<List<Integer>> fourSum(int[] nums, int target) {
// }
// 2. Dedup through results. It's much simpler than the above one.
/*
[1,0,-1,0,-2,2] - [-2,-1,0,0,1,2]
0
*/
public List<List<Integer>> fourSum(int[] nums, int target) {
if (nums == null || nums.length < 4) return new ArrayList<>();
List<List<Integer>> result = new ArrayList<>();
Arrays.sort(nums);
for (int i = 0; i < nums.length - 3; i++) {
for (int j = i + 1; j < nums.length - 2; j++) {
int lo = j + 1, hi = nums.length - 1, rest = target - nums[i] - nums[j];
while (lo < hi) {
if (nums[lo] + nums[hi] == rest) {
List<Integer> list = new ArrayList<>();
list.add(nums[i]);
list.add(nums[j]);
list.add(nums[lo]);
list.add(nums[hi]);
if (!result.contains(list)) {
result.add(list);
}
lo++;
hi--;
} else if (nums[lo] + nums[hi] < rest) {
lo++;
} else {
hi--;
}
}
}
}
return result;
}
}