[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].

 

這道題就是最基本的二分搜索法了，這是博主之前總結的LeetCode Binary Search Summary 二分搜索法小結的四種之中的第一類，也是最簡單的一類，寫法什么很模版啊，注意right的初始化值，還有while的循環條件，以及right的更新值，這三點不同的人可能會有不同的寫法，選一種自己最習慣的就好啦，參見代碼如下：

 

class Solution {
public:
    int search(vector<int>& nums, int target) {
        int left = 0, right = nums.size();
        while (left < right) {
            int mid = left + (right - left) / 2;
            if (nums[mid] == target) return mid;
            else if (nums[mid] < target) left = mid + 1;
            else right = mid;
        }
        return -1;
    }
};

 

類似題目：

Search in a Sorted Array of Unknown Size

 

參考資料：

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

 

LeetCode All in One 題目講解匯總(持續更新中...)