lintcode458- Last Position of Target- easy

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

Example

Given [1, 2, 2, 4, 5, 5].

For target = 2, return 2.

For target = 5, return 5.

For target = 6, return -1.

用二分法模板，缩范围时想一下怎么保证当前的数字不会被丢掉就好了。

其实就是因为这种要找最后一个第一个的题目才产生这种模板的。要最后一个所以逼着改指针不写= mid - 1写= mid，所以while条件不写start < end写start + 1 < end，所以最后要判断两个数而不是一个数。

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