[Leetcode]-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 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.

Hide Tags ：Array 。Two Pointers
题目26：输入一个已经拍好序的数组。当中有可能有反复数据。找出反复数据。栓出多余的反复数据。仅仅保留一个反复数据。

思路：记录当中的反复数据次数。比方当中有三个数据都是5，那么反复次数就为2。

将后面的数据依次填充到反复数据位置.

if(nums[i] == nums[i+1]) count++;
else nums[i+1-count] = nums[i+1]; 

题目27：本题和题26相似。27题要求将与给定数据val反复数据都栓出，不保留不论什么一个反复数据，方法相似，记录反复次数比26题多1个即可

if(nums[i] == val) count++;
else nums[i-count] = nums[i]; 

#include <stdlib.h>
#include <stdio.h>
//26: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 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.
  */
//思路一直累加反复的个数，数组第i 位= i - 反复的个数
int removeDuplicates(int* nums, int numsSize) {

    int i = 0;
    int count = 0;

    for(i=0;i<numsSize-1;i++)
    {
        if(nums[i] == nums[i+1])
        {
            count++;
        }
        else
        {
            nums[i+1-count] = nums[i+1];    
        }   

    }

    return numsSize-count;
}

//27 :Remove Element 
//Given an array and a value, remove all instances of that value in place and return the new length.
//The order of elements can be changed. It doesn't matter what you leave beyond the new length.

int removeElement(int* nums, int numsSize, int val) {
    int i = 0;
    int count = 0;
    for(i=0;i<numsSize;i++)
    {
        if(nums[i] == val)
        {
            count++;
        }
        else
        {
           nums[i-count] = nums[i]; 
        }
    }

    return numsSize-count;
}


int main()
{
    int n[10] = {1,2,3,3,3,4,5,5,6,7};
    int s[10] = {1,2,3,4,5,6,7,8,9,10};
    int r = removeDuplicates(s,10);
    printf("r = %d\n",r);
    int i = 0 ;
    for(i=0;i<r;i++)
    {
        printf("n[%d] = %d\n",i,n[i]);
    }
}

来源：https://www.cnblogs.com/gcczhongduan/p/5245679.html