[Leetcode 704]Binary Search

Description:

Given an array of integers nums which is sorted in ascending order, and an integer target, write a function to search target in nums. If target exists, then return its index. Otherwise, return -1.

You must write an algorithm with O(log n) runtime complexity.

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

C code solution:

int search(int* nums, int numsSize, int target){
int idx = -1;
int left = 0;
int right = numsSize-1;
while(left <= right){
int mid = (left + right) / 2;
if(nums[mid] < target){
left = mid+1;
}else if(nums[mid] > target){
right = mid-1;
}else{
idx = mid;
break;
}
}
return idx;
}

Explanation:

每次都選擇檢查中間的那個數。一開始設定檢查一整個array,設定left為左側index,right為右側index,算出mid為中間index。檢查mid是否為target。若否且比target小,代表target可能在右側,所以將left設成mid+1。若比target大,代表target可能在左側,所以將right設成mid-1。重複找每個區段中間那個數直到不符合left≤right。時間複雜度為O(logn)。