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Algorithm

Given a sorted array nums, remove the duplicates in-place such that each element appear only once and return the new length.

Do not allocate extra space for another array, you must do this by modifying the input array in-place with O(1) extra memory.
Example 1:

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Given nums = [1,1,2],

Your function should return length = 2, with the first two elements of nums being 1 and 2 respectively.

It doesn't matter what you leave beyond the returned length.

Example 2:

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Given nums = [0,0,1,1,1,2,2,3,3,4],

Your function should return length = 5, with the first five elements of nums being modified to 0, 1, 2, 3, and 4 respectively.

It doesn't matter what values are set beyond the returned length.

Clarification:

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Confused why the returned value is an integer but your answer is an array?

Note that the input array is passed in by reference, which means modification to the input array will be known to the caller as well.

Internally you can think of this:s

// nums is passed in by reference. (i.e., without making a copy)
int len = removeDuplicates(nums);

// any modification to nums in your function would be known by the caller.
// using the length returned by your function, it prints the first len elements.
for (int i = 0; i < len; i++) {
    print(nums[i]);
}

answer
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class Solution(object):
    def removeDuplicates(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        for i in range(len(nums)-1, 0, -1):
            if nums[i] == nums[i-1]:
                nums.pop(i)
        return len(nums)

Rewiew

Tips

rabbit mq启动UI管理界面
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rabbitmq-plugins enable rabbitmq_management
rabbitmq-server

启动rabbitmq_management插件，再启动rabbitmq-server即可。

rabbit mq用户管理

增加用户，设置权限

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rabbitmqctl add_user test test
rabbitmqctl set_user_tags test administrator
rabbitmqctl set_permissions -p / test ".*" ".*" ".*"

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