458. Last Position of Target [LintCode]

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.

要找last index，所以指针尽可能往右边靠：因此当中间数等于target的时候，是把left指针往右边移动；最后判断的两根指针的值的时候先判断right指针。

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