Last Position of Target

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Find the last position of a target number in a sorted array. Return -1 if target does not exist.

Example

Example 1:

Input: nums = [1,2,2,4,5,5], target = 2
Output: 2

Example 2:

Input: nums = [1,2,2,4,5,5], target = 6
Output: -1

思路1：用recursion写，自己写出来的，开心。

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) {if(nums == null || nums.length == 0) return -1;int start = 0; int end = nums.length-1;return findLastIndex(nums, start, end, target);}public int findLastIndex(int[] nums, int start, int end, int target) {if(start > end) {return -1;}int mid = start +(end-start) / 2;if(nums[mid] == target){if(mid == nums.length-1) return mid;if(nums[mid] != nums[mid+1]){return mid;} return findLastIndex(nums, mid+1, end, target);} else if(nums[mid] < target) {return findLastIndex(nums, mid+1, end, target);} else { // nums[mid] > target;return findLastIndex(nums, start, mid-1, target);}}
}

思路2： 用while循环写，按照九章的模板来写。

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) {if(nums == null || nums.length == 0) {return -1;}int start = 0; int end = nums.length -1;while(start + 1 < end) {int mid = start + (end - start) / 2;if(nums[mid] <= target) {start = mid;} else {// nums[mid] > target;end = mid;}}if(nums[end] == target){return end;}if(nums[start] == target){return start;}return -1;}
}

 

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