704. Binary Search - Easy
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 targetexists, then return its index, otherwise return -1.
Example 1:
Input:nums= [-1,0,3,5,9,12],target= 9 Output: 4 Explanation: 9 exists innumsand its index is 4
Example 2:
Input:nums= [-1,0,3,5,9,12],target= 2 Output: -1 Explanation: 2 does not exist innumsso return -1
Note:
- You may assume that all elements in
numsare unique. nwill be in the range[1, 10000].- The value of each element in
numswill be in the range[-9999, 9999].
classical binary search
time: O(logn), space: O(1)
class Solution { public int search(int[] nums, int target) { if(nums == null || nums.length == 0) { return -1; } int left = 0, right = nums.length - 1; 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 - 1; } } return -1; } }
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