last occurance
Given a target integer T and an integer array A sorted in ascending order, find the index of the last occurrence of T in A or return -1 if there is no such index.
Assumptions
-
There can be duplicate elements in the array.
Examples
- A = {1, 2, 3}, T = 2, return 1
- A = {1, 2, 3}, T = 4, return -1
- A = {1, 2, 2, 2, 3}, T = 2, return 3
Corner Cases
- What if A is null or A is array of zero length? We should return -1 in this case.
1 public int lastOccur(int[] array, int target) {
2 // Write your solution here
3 if(array == null || array.length == 0 ){
4 return -1 ;
5 }
6 int left = 0, right = array.length -1 ;
7 while(left + 1 <right){
8 int mid = left + (right - left)/2 ;
9 //因为找最后的，所以碰上也不扔着，带着往后面找
10 if(array[mid]<=target){
11 left = mid ;
12 } else{
13 right = mid ;
14 }
15 }
16 //post processing
17 if(array[right] == target){
18 return right ;
19 }
20 if(array[left] == target){
21 return left;
22 }
23 return -1 ;
24 }
