# Classical Binary Search

ID: 457; easy; 经典二分查找问题

## Solution 1 (Java)

``````public class Solution {
/**
* @param nums: An integer array sorted in ascending order
* @param target: An integer
* @return: An integer
*/
public int findPosition(int[] nums, int target) {
if (nums == null || nums.length == 0) {
return -1;
}
int left = 0;
int right = nums.length - 1;
while (left + 1 < right) {
int mid = left + (right - left) / 2;
if (nums[mid] == target) {
return mid;
} else if (nums[mid] < target) {
left = mid;
} else {
right = mid;
}
}
if (nums[left] == target) return left;
if (nums[right] == target) return right;
return -1;
}
}``````

### Notes

• The loop condition `left + 1 < right` will guarantee that we end up with two indices that are next to each other. Then, we return either of them at the end.

• The middle point `mid` is calculated using `left + (right - left) / 2` instead of `(left + right) / 2` in order to prevent integer overflow, which the addition may have while the subtraction will not.

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