[Lintcode] 14. First Position of Target_lintcodec++14. first position of target-优快云博客

本文链接：https://blog.youkuaiyun.com/siriusli1981/article/details/86763839

本文探讨了在已排序的数组中使用二分查找法，以O(logn)的时间复杂度找到目标数的第一个出现位置的算法。通过三个实例，详细解释了如何实现这一功能，并讨论了当数据量超过2^32时的挑战。

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14. First Position of Target

Description

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?

我的代码


class Solution:
    """
    @param nums: The integer array.
    @param target: Target to find.
    @return: The first position of target. Position starts from 0.
    """
    def binarySearch(self, nums, target):
        # write your code here
        l = 0
        r = len(nums)-1
        m = int((l+r)/2)
        while(l<=r):
            if nums[m]==target and ( m==0 or nums[m-1]<target):
                return m
            else:
                if nums[m]>=target:
                    r = m-1
                else:
                    l = m+1
                m = int((l+r)/2)
                
        return -1