Last Position of Target

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.

来源： http://www.lintcode.com/en/problem/first-position-of-target/

分析

找排序数组某个数字的右边界

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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;     } }

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