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].

 

A solution of o(logn). Do two runs of binary search for the index of largest element strictly less than target, and the index of largest element strictly less than target+1. The range is between this two indices. 

 1 public class Solution {
 2     public int[] searchRange(int[] A, int target) {
 3         // Start typing your Java solution below
 4         // DO NOT write main() function
 5         if(A==null || A.length==0)return null;
 6         int[] res = new int[2];
 7         res[0]=searchMaxLessThan(A,target,0,A.length-1);
 8         res[1]=searchMaxLessThan(A,target+1,0,A.length-1);
 9         if(res[0]==res[1]){
10             res[0]=-1;
11             res[1]=-1;
12         }else{
13             res[0]++;
14         }
15         return res;
16     }
17     
18     public int searchMaxLessThan(int[] A, int target, int start, int end){
19         
20         if(start==end) return A[start]<target?start:start-1;
21         if(start==end-1) return A[end]<target?end:(A[start]<target?start:start-1);
22         int mid = (start+end)/2;
23         if(A[mid]>=target){
24             end=mid-1;
25         }else{
26             start=mid;
27         }
28         return searchMaxLessThan(A,target,start,end);
29     }
30 }

 第二遍：

很巧妙的采用两次循环来做。

 1 public class Solution {
 2     public int[] searchRange(int[] A, int target) {
 3         // Start typing your Java solution below
 4         // DO NOT write main() function
 5         int start = 0;
 6         int end = A.length - 1;
 7         int l = -1;
 8         int r = -1;
 9         while(start <= end){
10             int mid = (start + end) / 2;
11             if(A[mid] == target) l = mid;
12             if(target <= A[mid]) end = mid - 1;
13             else start = mid + 1;
14         }
15         start = 0;
16         end = A.length - 1;
17         while(start <= end){
18             int mid = (start + end) / 2;
19             if(A[mid] == target) r = mid;
20             if(target >= A[mid]) start = mid + 1;
21             else end = mid - 1;
22         }
23         return new int[]{l, r};
24     }
25 }

 第三遍：

 1 public class Solution {
 2     public int[] searchRange(int[] A, int target) {
 3         int len = A.length;
 4         int srt = 0, end = len - 1;
 5         int left = -1, right = -1;
 6         while(srt <= end){
 7             int mid = (srt + end) / 2;
 8             if(A[mid] < target) srt = mid + 1;
 9             else if(A[mid] > target) end = mid - 1;
10             else{
11                 left = mid;
12                 end = mid - 1;
13             }
14         }
15         srt = 0; end = len - 1;
16         while(srt <= end){
17             int mid = (srt + end) / 2;
18             if(A[mid] < target) srt = mid + 1;
19             else if(A[mid] > target) end = mid - 1;
20             else{
21                 right = mid;
22                 srt = mid + 1;
23             }
24         }
25         return new int[]{left, right};
26     }
27 }

 

posted on 2013-10-12 14:08  Step-BY-Step  阅读(152)  评论(0)  编辑  收藏  举报

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