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 }
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 }
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 }
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 }
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 }
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 }
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 }
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 }
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 }
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
原文地址：https://www.cnblogs.com/lizzyluvcoding/p/9598874.html
时间： 2024-10-10 08:58:44