Search for a Range

Given a sorted array of integers, find the starting and ending position of a given target value.

Your algorithm's runtime complexity must be in the order of O(log n).

If the target is not found in the array, return [-1, -1].

For example,
Given [5, 7, 7, 8, 8, 10] and target value 8,
return [3, 4].

思想：二分递归的方法和二分查找，明确最后终止的位置

1. int search(int[] A,int target,int start,int end,boolean flag){
2. int res = 0;
3. if(flag) {
4. res = end;
5. } else {
6. res = start;
7. }
8. while(start<=end) {
9. int mid = start + (end-start)/2;
10. if(A[mid] == target) {
11. if(flag) {
12. if(start!=mid) {
13. res = search(A,target,start,mid,true); //注意退出条件
14. break; 
15. }
16. else {
17. res= start;
18. break;
19. }
20. } else {
21. if(mid+1>=end) {
22. if(mid==end) res = mid;
23. else {
24. if(A[end] == target) res = end;
25. else res = mid;
26. }
27. break;
28. } else {
29. res = search(A,target,mid,end,false);
30. break;
31. }
32. }
33. } else if(A[mid] <target){
34. start = mid + 1;
35. } else {
36. end = mid -1;
37. }
38. }
39. return res;
40. }
41. public int[] searchRange(int[] A, int target) {
42. int length = A.length;
43. int[] res = new int[2];
44. if(length == 0) {
45. res[0] = -1;
46. res[1] = -1;
47. return res;
48. }
49. int start = 0;
50. int end = length-1;
51. int sindex = -1;
52. int eindex = -1;
53. while(start <= end) {
54. int mid = start + (end - start)/2;
55. if(A[mid] == target) {
56. sindex = search(A,target,start,mid,true);
57. eindex = search(A,target,mid,end,false);
58. break;
59. } else if(A[mid] < target) {
60. start = mid + 1;
61. } else {
62. end = mid - 1;
63. }
64. }
65. res[0] = sindex;
66. res[1] = eindex;
67. return res;
68. }

改进无递归：

1. public int[] searchRange(int[] A, int target) {
2. int length = A.length;
3. int[] res = new int[2];
4. res[0] = -1;
5. res[1] = -1;
6. if(length == 0) {
7. return res;
8. }
9. int start = 0;
10. int end = length-1;
11. int sindex = -1;
12. int eindex = -1;
13. int find = -1;
14. while(start <= end) {
15. int mid = start + (end - start)/2;
16. if(A[mid] == target) {
17. find = mid;
18. break;
19. } else if(A[mid] < target) {
20. start = mid + 1;
21. } else {
22. end = mid - 1;
23. }
24. }
25. if(find == -1) {
26. return res;
27. }
29. start = 0;
30. sindex = find;
31. while(start < sindex) {
32. int mid = start + (sindex-start)/2;
33. if(A[mid] == target) {
34. sindex = mid;
35. } else {
36. start = mid + 1;
37. }
38. }
39. res[0] = sindex;
40. end = A.length - 1;
41. eindex = find;
42. while(eindex<end) {
43. int mid = eindex + (end - eindex)/2;
44. if(A[mid] == target) {
45. eindex = mid;
46. if(mid+1==end) {
47. if(A[end] == target) eindex = end;
48. break;
49. }
50. } else {
51. end = mid - 1;
52. }
53. }
54. res[1] = eindex;
55. return res;
56. }

左区间和右区间优化

1. start = 0;
2. sindex = find;
3. while(start <= sindex) {
4. int mid = start + (sindex-start)/2;
5. if(A[mid] == target) {
6. sindex = mid-1;
7. } else {
8. start = mid + 1;
9. }
10. }
11. res[0] = start;
12. end = A.length - 1;
13. eindex = find;
14. while(eindex<=end) {
15. int mid = eindex + (end - eindex)/2;
16. if(A[mid] == target) {
17. eindex = mid + 1;
18. } else {
19. end = mid - 1;
20. }
21. }
22. res[1] = end;
23. return res;