Lintcode：二分查找

给定一个排序的整数数组（升序）和一个要查找的整数target，用O(logn)的时间查找到target第一次出现的下标（从0开始），如果target不存在于数组中，返回-1。

在数组 [1, 2, 3, 3, 4, 5, 10] 中二分查找3，返回2。

python：

class Solution:    # @param nums: The integer array    # @param target: Target number to find    # @return the first position of target in nums, position start from 0    def binarySearch(self, nums, target):        # write your code here        low = 0        high = len(nums) - 1        while low <= high:            middle = (high + low) // 2            if nums[middle] < target:                low = middle + 1            elif nums[middle] > target:                high = middle - 1            else:                while nums[middle] == nums[middle - 1]:                    middle -= 1                return middle        return -1

C++：

class Solution {public:    /**     * @param nums: The integer array.     * @param target: Target number to find.     * @return: The first position of target. Position starts from 0.      */    int binarySearch(vector<int> &array, int target) {        // write your code here        if (array.size() == 1)        {            if (array[0] == target)                return 0;        }        int start = 0;        int end = array.size();        int middle = end / 2;        while (array[middle] != target)        {            if (array[middle] > target)            {                end = middle - 1;            }            else            {                start = middle + 1;            }            middle = (start + end) / 2;            if (array[middle] < target && array[middle + 1] > target)                return -1;        }        while (array[middle] == array[middle - 1])        {            middle--;        }        return middle;    }};