Lintcode 14 first position of target

最新推荐文章于 2022-01-04 17:50:05 发布

原创最新推荐文章于 2022-01-04 17:50:05 发布 · 252 阅读

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题目：

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

If the array is [1, 2, 3, 3, 4, 5, 10], for given target 3, return 2.

解析：这个题考的是二分查找，但是有一点不同的是要求返回符合的第一个数在数组中的位置

代码如下：

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) {
// write your code here
int mid = 0;
int start = 0;
int end = nums.length-1;
while(start<end)
{
mid = (end + start)/2;
if(nums[mid]>target)
end = mid - 1;
if(nums[mid]<target)
start = mid + 1;
else if(nums[mid]==target)
end = mid;
}
if(nums[end]==target)
return end;
return -1;
}
}