第二周作业1(LeetCode27)

1. 题目描述(LeetCode27)

Given an array and a value, remove all instances of that value in place and return the new length.

Do not allocate extra space for another array, you must do this in place with constant memory.

The order of elements can be changed. It doesn’t matter what you leave beyond the new length.

Example:

Given input array nums = [3,2,2,3], val = 3

Your function should return length = 2, with the first two elements of nums being 2.

2. 解决思路

此题的关键是采用何种方式来删除指定的元素以及对剩下元素的处理。认真观察删除操作的性质会发现，删除一个元素不过就是把他后面的元素向前移动一位，那删除两个就移动两位，三个就是三位。。。以此类推。这样的时间复杂度就为O(n)。

3. 完整代码

#include <stdio.h>
#include <malloc.h>

//数组S, len数组长度，要删除的数val,返回新的数组长度
int RemoveElement(int* S, int len, int val);

int main()
{
int* S = NULL;  //存储数组
int len = 0;
int val = 0;
printf("请输入数组的长度：\n");
scanf("%d", &len);
while (len<= 0)
{
printf("长度必须大于0，请重新输入：\n");
scanf("%d", &len);
}
S = (int*)malloc(sizeof(int) * len);  //数组分配内存

printf("请输入数组的每一位数：\n");
for (int i = 0; i<len; ++i)
{
scanf("%d", &S[i]);
}

printf("请输入要删除的数：\n");
scanf("%d", &val);

//调用封装好的功能函数
int result = RemoveElement(S,len, val);

//输出结果
printf("新的数组长度为: %d\n", result);

printf("新的数组元素:\n");
for (int i=0; i<result ; ++i)
{
printf("%d ", S
9720
[i]);
}

scanf("%d", &len);  //加个输入让窗口停下来

free(S);  //释放内存
return 0;
}

int RemoveElement(int* S,int len, int val)
{

for (int i = 0; i< len ; ++i)
{
if (S[i] == val)
{

for (int j = i; j < len - 1; ++j)  //将后面的元素前移
{
S[j] = S[j+1];
}
len -= 1; //删除后数组长度减1
}
}

return len;

}