``#### Last Position of Target

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

Given[1, 2, 2, 4, 5, 5].

For target =2, return 2.

For target =5, return 5.

For target =6, return -1.

Solution: Binary Search. One for target, one for last position index.

public int lastPosition(int[] nums, int target) {
    if(nums == null || nums.length == 0)
        return -1;
    int start = 0, 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;
}