[LeetCode] 26. Remove Duplicates from Sorted Array ☆(从有序数组中删除重复项)

[LeetCode] Remove Duplicates from Sorted Array 有序数组中去除重复项

描述

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]);
}

给定一个有序数组，你需要原地删除其中的重复内容，使每个元素只出现一次,并返回新的长度。 
不要另外定义一个数组，您必须通过用 O(1) 额外内存原地修改输入的数组来做到这一点。 
示例：

给定数组: nums = [1,1,2],
你的函数应该返回新长度 2, 并且原数组nums的前两个元素必须是1和2
不需要理会新的数组长度后面的元素

解析

我们使用快慢指针来记录遍历的坐标，最开始时两个指针都指向第一个数字，如果两个指针指的数字相同，则快指针向前走一步，如果不同，则两个指针都向前走一步，这样当快指针走完整个数组后，慢指针当前的坐标加1就是数组中不同数字的个数。

代码

class Solution {
    public int removeDuplicates(int[] nums) {
        if (nums == null || nums.length <= 0) {
            return 0;
        }
        if (nums.length < 2) {
            return 1;
        }
        int pre = 0, cur = 0, n = nums.length;
        while (cur < n) {
            if (nums[pre] == nums[cur])
                ++cur;
            else
                nums[++pre] = nums[cur++];
        }
        return pre + 1;
    }
}

我们也可以用for循环来写，这里的j就是上面解法中的pre，i就是cur，所以本质上都是一样的，参见代码如下：

class Solution {
public:
    int removeDuplicates(vector<int>& nums) {
        if (nums.empty()) return 0;
        int j = 0, n = nums.size();
        for (int i = 0; i < n; ++i) {
            if (nums[i] != nums[j]) nums[++j] = nums[i];
        }
        return j + 1;
    }
};