[LeetCode] Binary Search 二分搜索法

Given a sorted (in ascending order) integer array nums of n elements and a target value, write a function to search target in nums. If target exists, then return its index, otherwise return -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:

1. You may assume that all elements in nums are unique.
2. n will be in the range [1, 10000].
3. The value of each element in nums will be in the range [-9999, 9999].

这道题就是最基本的二分搜索法了，这是博主之前总结的LeetCode Binary Search Summary 二分搜索法小结的四种之中的第一类，也是最简单的一类，写法什么很模版啊，注意right的初始化值，还有while的循环条件，以及right的更新值，这三点不同的人可能会有不同的写法，选一种自己最习惯的就好啦，参见代码如下：

class Solution {
public:
    int search(vector<int>& nums, int target) {
        int left = , right = nums.size();
        while (left < right) {
            int mid = left + (right - left) / ;
            if (nums[mid] == target) return mid;
            else if (nums[mid] < target) left = mid + ;
            else right = mid;
        }
        return -;
    }
};

类似题目：

Search in a Sorted Array of Unknown Size

参考资料：

https://leetcode.com/problems/binary-search

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