Leetcode_remove-duplicates-from-sorted-array-优快云博客

本文介绍了一种方法，在不使用额外空间的情况下，去除有序数组中的重复元素，并返回新的数组长度。提供了两种不同的实现方式，适用于不同场景。

地址：http://oj.leetcode.com/problems/remove-duplicates-from-sorted-array/

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 A = [1,1,2],

Your function should return length = 2, and A is now [1,2].

思路：在循环内，记下当前已发现的重复个数diff，应将当前i的值A[i] 赋给 A[i-diff]. 返回的个数便是n-diff

参考代码：

class Solution {
public:
    int removeDuplicates(int A[], int n) {
        int diff = 0;
        for(int i = 0; i < n-1; ++i)
        {
            if(A[i]==A[i+1])
            {
                while(i < n-1 && A[i]==A[i+1])
                {
                    ++diff;
                    ++i;
                }
            }
            A[i-diff] = A[i];
        }
        A[n-1-diff] = A[n-1];
        return n-diff;
    }
};

SECOND TRIAL, 发现前后两次的解法差距还是挺大的

class Solution {
public:
    int removeDuplicates(int A[], int n) {
        if(n<=1)
            return n;
        int cnt = 1;
        for(int i = 1; i<n; ++i)
        {
            if(A[i]!=A[cnt-1])
            {
                A[cnt] = A[i];
                ++cnt;
            }
        }
        return cnt;
    }
};

Pyton version:

class Solution:
    # @param a list of integers
    # @return an integer
    def removeDuplicates(self, A):
        if len(A) <= 1:
            return len(A)
        cnt = 1
        for i in range(1, len(A)):
            if A[i] != A[cnt-1]:
                A[cnt] = A[i]
                cnt += 1
        return cnt