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For a given sorted array (ascending order) and atarget
number, find the first index of this number inO(log n)
time complexity.
If the target number does not exist in the array, return -1
.
If the array is [1, 2, 3, 3, 4, 5, 10]
, for given target 3
, return2
.
If the count of numbers is bigger than 2^32, can your code work properly?
class Solution { /** * @param nums: The integer array. * @param target: Target to find. * @return: The first position of target. Position starts from 0. */ public int binarySearch(int[] nums, int target) { //write your code here if(nums == null || nums.length == 0) return 0; int left = 0; int right = nums.length - 1; while(left < right){ int mid = left + (right - left) / 2; if(nums[mid] < target) left = mid + 1; else if(nums[mid] > target) right = mid - 1; else right = mid; } if(nums[left] == target) return left; else return -1; } }
lintcode-easy-First Position of Target
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原文地址:http://www.cnblogs.com/goblinengineer/p/5213842.html