704. Binary Search
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
Constraints:
- 1 <= nums.length <= 10^4
- -10^4 < nums[i], target < 10^4
- All the integers in nums are unique.
- nums is sorted in ascending order.
class Solution {
//nums用到的数组。 target所选择的索引目标
public int search(int[] nums, int target) {
//对nums数组进行遍历
for(int i =0;i < nums.length;i++){
//如果nums遍历之中的索引i的值 正好等于 target
if(nums[i] == target){
//那么返回i
return i;
}
}
//否则返回-1
return -1;
}
}