> For the complete documentation index, see [llms.txt](https://rachel2011.gitbook.io/leetcode_cc150/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://rachel2011.gitbook.io/leetcode_cc150/binary-search/last-position-of-target.md).

# Last Position of Target

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.

1. 当找到与target相等的值时，继续向右寻找（start＝mid），因为题目要求寻找最后一个index
2. 循环结束判断时，需要先判断A\[end],然后再判断A\[start]

```
class Solution:
    # @param {int[]} A an integer array sorted in ascending order
    # @param {int} target an integer
    # @return {int} an integer
    def lastPosition(self, A, target):
        # Write your code here
        if len(A)==0:
            return -1
        start = 0
        end = len(A)-1
        while start+1<end:
            mid = (start+end)/2
            if target >= A[mid]:
                start = mid
            else:
                end = mid
        if target == A[end]:
            return end
        elif target == A[start]:
            return start
        return -1
```