-
Notifications
You must be signed in to change notification settings - Fork 1
Expand file tree
/
Copy pathtest31.java
More file actions
76 lines (71 loc) · 2.16 KB
/
test31.java
File metadata and controls
76 lines (71 loc) · 2.16 KB
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
package byteDance;
import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
/**
* Created by lizeyang on 2020/5/15.
* 递增数组，找出和为k的数对
*/
public class test31 {
//双指针，时间复杂度O(n)
public static int[][] test(int[] nums, int target) {
if (nums == null || nums.length == 0) {
return new int[][]{};
}
List<int[]> list = new ArrayList<>();
int l = 0, r = nums.length - 1;
while (l < r) {
if (nums[l] + nums[r] == target) {
int[] res = new int[2];
res[0] = nums[l];
res[1] = nums[r];
list.add(res);
l++;
r--;
} else if (nums[l] + nums[r] < target) {
l++;
} else {
r--;
}
}
int[][] res = new int[list.size()][];
for (int i = 0; i < list.size(); i++) {
res[i] = list.get(i);
}
return res;
}
//升级：和为k的子数组，如nuns = {1,1,1},k=2;输出[1,1][1,1]
//利用sum-k这样的查补方式找到map里面对应sum-k的value值
public static int subArraySum(int[] nums, int k) {
if (nums == null || nums.length == 0) {
return 0;
}
HashMap<Integer, Integer> map = new HashMap<>();
map.put(0, 1);
int sum = 0;
int count = 0;
for (int i = 0; i < nums.length; i++) {
sum += nums[i];
if (map.containsKey(sum - k)) {
count += map.get(sum - k);
}
if (!map.containsKey(sum)) {
map.put(sum, 1);
} else {
map.put(sum, map.get(sum) + 1);
}
}
return count;
}
public static void main(String[] args) {
int[] nums = {1, 2, 4, 6, 7, 8};
int[][] res = test(nums, 9);
for (int i = 0; i < res.length; i++) {
for (int j = 0; j < res[0].length; j++) {
System.out.print(res[i][j] + " ");
}
System.out.println();
}
System.out.println(subArraySum(nums, 6));
}
}