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.

解题：找target最后出现的数组index，用二分发来解，当mid值大于target，我们更新hi为mid，反之当mid值小于等于target，我们更新lo为mid。

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

        if (nums[hi] == target) {
            return hi;
        }
        if (nums[lo] == target) {
            return lo;
        }
        return -1;
    }
}