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Classical Binary Search.java
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62 lines (52 loc) · 1.25 KB
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E
while: start + 1 < end
mid = start + (end - start) / 2;
末尾double check start, end.
```
/*
Find any position of a target number in a sorted array.
Return -1 if target does not exist.
Example
Given [1, 2, 2, 4, 5, 5].
For target = 2, return 1 or 2.
For target = 5, return 4 or 5.
For target = 6, return -1.
Challenge
O(logn) time
Tags Expand
Binary Search
*/
/*
Thoughts: classic
start,mid,end
*/
public class Solution {
/**
* @param A an integer array sorted in ascending order
* @param target an integer
* @return an integer
*/
public int findPosition(int[] A, int target) {
if (A == null || A.length == 0) {
return -1;
}
int start = 0;
int end = A.length - 1;
int mid;
while(start + 1 < end) {
mid = start + (end - start) / 2;
if (target == A[mid]) {
return mid;
} else if (target > A[mid]) {
start = mid;
} else {
end = mid;
}
}//end while
if (A[start] == target || A[end] == target) {
return A[start] == target ? start : end;
}
return -1;
}
}
```