Question

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]); }

Solution

class Solution {
 public:
  int removeDuplicates(vector<int>& nums) {
    if (nums.size() <= 1) return nums.size();
    int pre = 0, cur = 1, ctr = 1;
    while (cur < nums.size()) {
      if (nums[pre] == nums[cur] && ctr == 0)
        cur++;
      else {
        if (nums[pre] == nums[cur])
          ctr = 0;
        else {
          nums[++pre] = nums[cur++];
          ctr = 1;
        }
      }
    }
    return pre + 1;
  }
};

上面的解决方案有这样几个问题：

• nums.size() <= 1 不应该将长度为1的情况归入和0一样的处理方式，这样不够清晰明了。

对上面的解答思路进行优化：

Solution 2

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

Solution 3

使用了STL里面的函数来快速的完成：

class Solution {
 public:
  int removeDuplicates(vector<int>& nums) { return distance(nums.begin(), unique(nums.begin(), nums.end())); }
};

The Template for the remove the duplicates.

// 删除重复的元素
templaate<typename InIt, typename OutIt>
OutIt removeDuplicates(InIt first, InIt last, OutIt output) {
  while(first != last) {
    *output++ = *first;
    first = upper_bound(first, last, *first);
  }
  return output;
}