二分二刷-20190205

Posted 2021-02-08 lizzyluvcoding

二分法的时间复杂度：logN

 

算法面试复杂度：

1.O(1) 不出现

2.o(logN) 二分法

3.o(n^1/2) 分解质因数

4.o(n) 高频

5.o(nlogn) 一般排序

6.O(N^2)数组 枚举 动态规划

7.O(N^3) 数组 枚举 动态规划

8.O(2^n) 与组合有关搜索

9.O(n!) 与排列有关的搜索

 

https://leetcode.com/problems/find-first-and-last-position-of-element-in-sorted-array/solution/

 1 class Solution {
 2     public int[] searchRange(int[] nums, int target) {
 3         int[] res = new int[2];
 4         res[0] =-1;
 5         res[1] = -1;
 6         if(nums==null || nums.length==0){
 7             return res;
 8         }
 9         
10         
11         int start =0;
12         int end = nums.length-1;
13         while(start +1 <end){
14             int mid = start +(end-start)/2;
15             if(nums[mid]==target){
16                 end = mid;
17             }else if(nums[mid]>target){
18                 end = mid;
19             }else{
20                 start = mid;
21             }
22         }
23         if(nums[end]==target){
24             res[0] = end;
25         }
26         if(nums[start]==target){
27             res[0] = start;
28         }
29         
30         start =0;
31         end = nums.length-1;
32         
33         while(start +1 <end){
34             int mid = start +(end-start)/2;
35             if(nums[mid]==target){
36                 start = mid;
37             }else if(nums[mid]>target){
38                 end = mid;
39             }else{
40                 start = mid;
41             }
42         }
43         if(nums[start]==target){
44             res[1] = start;
45         }
46         if(nums[end]==target){
47             res[1] = end;
48         }
49         
50         return res;
51     }
52 }

View Code

 

https://leetcode.com/problems/first-bad-version/submissions/

 1 /* The isBadVersion API is defined in the parent class VersionControl.
 2       boolean isBadVersion(int version); */
 3 
 4 public class Solution extends VersionControl {
 5     public int firstBadVersion(int n) {
 6         if(n<1){
 7             return 0;
 8         }
 9         
10         int start =1;
11         int end = n;
12         
13         while(start + 1< end){
14             int mid = start +(end-start)/2;
15             if(isBadVersion(mid)){
16                 end = mid;
17             }else{
18                 start = mid;
19             }
20         }
21         
22         if(isBadVersion(start)){
23             return start;
24         }
25         if(isBadVersion(end)){
26             return end;
27         }
28         return 0;
29     }
30 }

View Code

https://www.lintcode.com/problem/search-in-a-big-sorted-array/description

 1 public class Solution {
 2     /*
 3      * @param reader: An instance of ArrayReader.
 4      * @param target: An integer
 5      * @return: An integer which is the first index of target.
 6      */
 7     public int searchBigSortedArray(ArrayReader reader, int target) {
 8         // write your code here
 9         if(reader.get(0)==target){
10             return 0;
11         }
12         int start = 1;
13         while(reader.get(start)<target){
14             start = start*2;
15         }
16         
17         int end = start;
18         start =1;
19         while(start+1<end){
20             int mid = start +(end-start)/2;
21             if(reader.get(mid)==target){
22                 end = mid;
23             }else if(reader.get(mid)>target){
24                 end = mid;
25             }else{
26                 start = mid;
27             }
28         }
29         
30         if(reader.get(start)==target){
31             return start;
32         }
33         if(reader.get(end) == target){
34             return end;
35         }
36         return -1;
37     }
38 }

View Code