For a given sorted array (ascending order) and a target number, find the first index of this number in O(log n) time complexity.
If the target number does not exist in the array, return -1.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 | public class Solution { /** * @param nums: An integer array sorted in ascending order * @param target: An integer * @return an integer */ public int lastPosition(int[] nums, int target) { // Write your code here if(nums == null || nums.length == 0) return -1; int left = 0, right = nums.length -1, mid; while(left < right){ mid = left + (right - left + 1) / 2;//insure when right is 1 bigger than left, mid eaqual to right if(nums[mid] > target){ right = mid - 1; } else{ left = mid; } } if(nums[right] == target) return right; else return -1; }} |
原文：http://www.cnblogs.com/zhxshseu/p/9fb92aa37069359c9204082cbf5d9632.html