[LeetCode] Binary Search [Beat 100%]
阿新 • • 發佈:2018-12-10
Problem
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 = 9Output: 4Explanation: 9 exists in nums and its index is 4
Example 2:
Input: nums = [-1,0,3,5,9,12], target = 2Output: -1Explanation: 2 does not exist in nums so return -1
Note:
You may assume that all elements in nums are unique.n will be in the range [1, 10000].The value of each element in nums will be in the range [-9999, 9999].
Beat 100%的要點:極簡主義的引數名,不考慮溢位的中點初始化
Solution
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)>>>1; if (nums[m] == target) return m; else if (nums[m] > target) r = m-1; else l = m+1; } return -1; } }