LeetCode 26 Remove Duplicates from Sorted Array (C,C++,Java,Python)

Problem:

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.

Solution:

由于是有序数组,所以当第i个元素和第i-1个元素相等的时候第i个元素一定是重复元素,采用两个游标,一个在左边,代表不重复的数组,一个在右边,代表原数组,当有不重复元素的时候,将这个元素赋值到不重复数组的顶部,并将左游标右移一位

题目大意:

给一个排好序的数组,要求移除所有重复元素,并且不能使用额外的存储空间。

Java源代码(337ms):

public class Solution {
    public int removeDuplicates(int[] nums) {
        int size=0,len=nums.length;
        for(int i=0;i<len;i++){
            if(i==0 || nums[i]!=nums[i-1]){
                nums[size++]=nums[i];
            }
        }
        return size;
    }
}

C语言源代码(30ms):

int removeDuplicates(int* nums, int numsSize) {
    int i,size=0;
    for(i=0;i<numsSize;i++){
        if(!(i>0 && nums[i]==nums[i-1])){
            nums[size]=nums[i];
            size++;
        }
    }
    return size;
}

C++源代码(38ms):

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

Python源代码(94ms):

class Solution:
    # @param {integer[]} nums
    # @return {integer}
    def removeDuplicates(self, nums):
        size=0;length=len(nums)
        for i in range(length):
            if i==0 or nums[i]!=nums[i-1]:nums[size]=nums[i];size+=1
        return size
时间: 05-11

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