LeetCode26. Remove Duplicates from Sorted Array（模拟）

Easy
Accepted：504,523
Submissions：1,286,129

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:

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:

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:

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:

// 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]);
}

链接

https://leetcode.com/problems/remove-duplicates-from-sorted-array/

题意

对于一个有序数组，去除其中的重复元素，返回新数组的长度。保证不使用额外的辅助空间，空间复杂度$O(1)$。

题解

题目说了是有序，所以就很好操作了。只需判断当前元素与前一个元素是否相等即可。

代码

Runtime: 16 ms, faster than 99.06% of C++ online submissions for Remove Duplicates from Sorted Array.

class Solution {
public:
    int removeDuplicates(vector<int>& nums) {
        int len = nums.size();
        int cnt = (len == 0 ? 0 : 1); // 考虑nums为空的情况
        for (int i = 1; i < len; ++i)
        {
            if (nums[i] != nums[i-1])
                nums[cnt++] = nums[i];
        }
        return cnt;
    }
};

切记这种写法需要考虑数组为空的情况，吃了一发runtime error。

The end.
2019年2月1日星期五