试题
For a given sorted array (ascending order) and a target number, find the first index of this number in O(log n) time complexity.
If the target number does not exist in the array, return -1.
Example
Example 1:
Input: [1,4,4,5,7,7,8,9,9,10],1
Output: 0
Explanation:
the first index of 1 is 0.
Example 2:
Input: [1, 2, 3, 3, 4, 5, 10],3
Output: 2
Explanation:
the first index of 3 is 2.
Example 3:
Input: [1, 2, 3, 3, 4, 5, 10],6
Output: -1
Explanation:
Not exist 6 in array.
Challenge
If the count of numbers is bigger than 2^32, can your code work properly?
代码
1、如果相等的话,要查找的值将会在左区间,并且有可能就是mid所在位置。
2、循环终止条件,对于重复的数最后left和right都会定位到第一个出现的位置,这是如果我们以left==right作为循环终止条件就会陷入死循环。
3、对于返回值,我们要判断最终查找位置的值是否等于target。
public class Solution {
/**
* @param nums: The integer array.
* @param target: Target to find.
* @return: The first position of target. Position starts from 0.
*/
public int binarySearch(int[] nums, int target) {
int left=0, right=nums.length-1;
while(left<right){
int mid = left+(right-left)/2;
if(nums[mid]==target){
right = mid;
}else if(nums[mid]>target){
right = mid-1;
}else if(nums[mid]<target){
left = mid+1;
}
}
if(nums[left]==target)
return left;
else
return -1;
}
}