704. Binary Search - Easy
Posted fatttcat
tags:
篇首语:本文由小常识网(cha138.com)小编为大家整理,主要介绍了704. Binary Search - Easy相关的知识,希望对你有一定的参考价值。
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 innums
and its index is 4
Example 2:
Input:nums
= [-1,0,3,5,9,12],target
= 2 Output: -1 Explanation: 2 does not exist innums
so return -1
classical binary search
time: O(log(n)), space: O(1)
class Solution { public int search(int[] nums, int target) { if(nums == null || nums.length == 0) return -1; int l = 0, r = nums.length - 1; while(l <= r) { int m = l + (r - l) / 2; if(nums[m] == target) return m; else if(nums[m] > target) r = m - 1; else l = m + 1; } return -1; } }
以上是关于704. Binary Search - Easy的主要内容,如果未能解决你的问题,请参考以下文章