【LeetCode】026. Remove Duplicates from Sorted Array

题目:

Given a sorted array, 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 in place with constant memory.

For example,
Given input array 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 new length.

题解：

　　感觉这个题是真没什么好解析的。。。

Solution 1()

class Solution {
public:
    int removeDuplicates(vector<int>& nums) {
        int n = nums.size();
        if(n < 2)
            return n;
        int index = 0;
        for(int i = 1; i < n; ++i)
            if(nums[index] != nums[i])
                nums[++index] = nums[i];
        return index + 1;
    }
};

Solution 2()

class Solution {
public:
    int removeDuplicates(vector<int>& nums) {
         int n = nums.size();
         int count = 0;
         for(int i = 1; i < n; i++){
            if(A[i] == A[i-1]) count++;
            else A[i-count] = A[i];
        }
        return n-count;
    }
};

 Solution 3 

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

为了避免多余的复制操作，可以扩展使用，比如类数组，当数组没有重复对象时，无需复制操作。而之前的两个解答将会依次复制。