标签：xpl uniq 解法 http val style try col entry
问题：
二分查找，给定一个已排序的数组，和一个目标值target
在该数组中找到target的index返回，若没找到，则返回-1。
Example 1: Input: nums = [-1,0,3,5,9,12], target = 9 Output: 4 Explanation: 9 exists in nums and its index is 4 Example 2: Input: nums = [-1,0,3,5,9,12], target = 2 Output: -1 Explanation: 2 does not exist in nums so return -1 Note: You may assume that all elements in nums are unique. n will be in the range [1, 10000]. The value of each element in nums will be in the range [-9999, 9999].
解法：二分查找(Binary Search)
参考 69. Sqrt(x) 的解释部分。
代码参考：
1 class Solution { 2 public: 3 int search(vector<int>& nums, int target) { 4 int l = 0, r = nums.size(); 5 while(l<r) { 6 int m = l+(r-l)/2; 7 if(nums[m] == target) return m; 8 if(nums[m] > target) { 9 r = m; 10 } else { 11 l = m+1; 12 } 13 } 14 return -1; 15 } 16 };
标签：xpl uniq 解法 http val style try col entry
原文地址：https://www.cnblogs.com/habibah-chang/p/13489106.html
