二分查找 - 20180906

时间：2018-09-07 18:01:01 阅读：184 评论：0 收藏：0 [点我收藏+]

14. First Position of Target

基本题

 1     public int binarySearch(int[] nums, int target) {
 2         // write your code here
 3         if(nums==null || nums.length==0){
 4             return -1;
 5         }
 6         
 7         int start =0;
 8         int end = nums.length-1;
 9         while(start+1<end){
10             int mid = start +(end-start)/2;
11             if(nums[mid]==target){
12                 end = mid;
13             }
14             else if(nums[mid]<target){
15                 start =mid;
16             }else{
17                 end =mid;
18             }
19         }
20         
21         if(nums[start]==target){
22             return start;
23         }
24         
25         if(nums[end]==target){
26             return end;
27         }
28         
29         return -1;
30     }

View Code

74. First Bad Version

14的变种

 1     public int findFirstBadVersion(int n) {
 2         // write your code here
 3         
 4         
 5         if(n<1){
 6             return -1;
 7         }
 8         
 9         int start = 1;
10         int end = n;
11         while(start+1<end){
12             int mid = start+(end-start)/2;
13             if(SVNRepo.isBadVersion(mid)){
14                 end = mid;
15             }
16             else {
17                 start = mid;
18             }
19         }
20         
21         if(SVNRepo.isBadVersion(start)){
22             return start;
23         }
24         
25         if(SVNRepo.isBadVersion(end)){
26             return end; 
27         }
28         
29         return -1;
30     }

View Code

458. Last Position of Target (同14，基本题)

 1     public int lastPosition(int[] nums, int target) {
 2         // write your code here
 3         if(nums==null || nums.length==0){
 4             return -1;
 5         }
 6         
 7         int start =0;
 8         int end = nums.length-1;
 9         while(start+1<end){
10             int mid = start +(end-start)/2;
11             if(nums[mid]==target){
12                 start = mid;
13             }
14             else if(nums[mid]<target){
15                 start =mid;
16             }else{
17                 end =mid;
18             }
19         }
20         
21         if(nums[end]==target){
22             return end;
23         }
24         
25         if(nums[start]==target){
26             return start;
27         }
28         
29         return -1;
30     }

View Code

447. Search in a Big Sorted Array

倍增的思想
1.数组扩容
2.网络爬虫

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

View Code

159. Find Minimum in Rotated Sorted Array

 1     public int findMin(int[] nums) {
 2         // write your code here
 3         if(nums==null || nums.length==0){
 4             return -1;
 5         }
 6         
 7         int start =0;
 8         int end = nums.length-1;
 9         while(start+1<end){
10             int mid = start + (end-start)/2;
11 
12             if(nums[mid]<nums[nums.length-1]){
13                 end = mid;
14             }
15              //注意判断条件不要写成nums[mid]>=nums[0],这种情况对于[1,2,3,4]这种未翻转的case会fail
16             else if(nums[mid]>=nums[nums.length-1]){
17                 start =mid;
18             }
19         }
20         
21         if(nums[start]<nums[end]){
22             return nums[start];
23         }else{
24             return nums[end];
25         }
26     }

View Code

585. Maximum Number in Mountain Sequence

此题需再次做，边界条件设置有问题提交一直不过

 1     public int mountainSequence(int[] nums) {
 2         // write your code here
 3         if(nums==null || nums.length==0){
 4             return -1;
 5         }
 6 
 7         int start = 0;
 8         int end = nums.length-1;
 9     
10         
11         while(start+1<end){
12             int mid = start+(end-start)/2;
13             if(nums[mid]-nums[mid+1]>0){
14                 end = mid;
15             }
16             else {
17                 start =mid;
18             }
19         }
20         
21         if(nums[start]<nums[end]){
22             return nums[end];
23         }else{
24             return nums[start];
25         }
26     }

View Code

75. Find Peak Element

和585其实一样

 1     public int findPeak(int[] A) {
 2         // write your code here
 3         
 4         int start =0;
 5         int end = A.length-2;
 6         
 7         while(start+1<end){
 8             int mid = start +(end-start)/2;
 9             if(A[mid]>A[mid+1]){
10                 end = mid;
11             }else{
12                 start =mid;
13             }
14         }
15         
16         return A[start]<A[end]?end:start;
17         
18     }

View Code

62. Search in Rotated Sorted Array

此题稍微复杂一点，分析一下几种情况再写，是经典题

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

View Code

28. Search a 2D Matrix

防出错，用了两次二分，其实也可以用一次二分解决（二刷时试试）

 1     public boolean searchMatrix(int[][] matrix, int target) {
 2         // write your code here
 3         
 4         if(matrix==null || matrix.length==0){
 5             return false;
 6         }
 7         
 8         //如何获取矩阵的行列
 9         int r = matrix.length;
10         int c = matrix[0].length;
11         
12         //先找出在哪一行
13         int start = 0;
14         int end = r-1;
15         while(start+1<end){
16             int mid = start + (end-start)/2;
17             if(matrix[mid][0]==target){
18                 return true;
19             }
20             else if(matrix[mid][0]>target){
21                 end = mid;
22             }
23             else{
24                 start = mid;
25             }
26         }
27         
28         int m = (matrix[end][0]<=target)?end:start;
29         
30         int mstart =0;
31         int mend = c-1;
32         while(mstart+1<mend){
33             int mmid = mstart +(mend-mstart)/2;
34             if(matrix[m][mmid]==target){
35                 return true;
36             }
37             else if(matrix[m][mmid]>target){
38                 mend = mmid;
39             }else{
40                 mstart = mmid;
41             }
42         }
43         
44         if(matrix[m][mstart]==target){
45             return true;
46         }
47         if(matrix[m][mend]==target){
48             return true;
49         }
50         
51         return false;
52     }

View Code

一刷主要刷基本题，二刷时可以考虑刷medium和hard，可以按高频次来刷

https://www.lintcode.com/problem/?tag=binary-search

二分查找 - 20180906

标签：seq ide medium star none 翻转 tps 情况注意

原文地址：https://www.cnblogs.com/lizzyluvcoding/p/9598874.html

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