To prove safety Offer --- 5 --- face questions replace spaces .md

To prove safety Offer- face questions 5 --- replace spaces

1 Introduction

Implement a function, each space in the replacement string as "20%." For example, type "Hello World", the output "Hello% 20World".

2, problem solution

Not recommended Method 1: traverse front to back, every encounter spaces and replace all the characters back to move backward.
Time complexity: O (n ^ 2)

Recommended Method 2: traversal from back to front, the detailed process of looking at the code.
Time complexity: O (n)

///arr是要修改的字符串，length表示这个字符串的最大长度
void ChangeSpace(char arr[], int length)
{
    if (arr == nullptr || length<=0)
    {
        return;
    }

    int i = 0;
    //字符串实际长度
    int arrLength = 0;
    //空格个数
    int spaceNum = 0;
    while (arr[i] != '\0')
    {
        if (arr[i] == ' ')
            spaceNum++;

        arrLength++;
        i++;
    }

    //新字符串需要的长度
    int newArrLength = arrLength + spaceNum * 2;
    if (newArrLength > length)
        return;

    //从后向前依次比较替换
    while (arrLength >= 0 && newArrLength > arrLength)
    {
        if (arr[arrLength] == ' ')
        {
            arr[newArrLength] = '0';
            arr[newArrLength-1] = '2';
            arr[newArrLength-2] = '%';
            arrLength--;
            newArrLength-=3;
        }
        else {
            arr[newArrLength] = arr[arrLength];
            arrLength--;
            newArrLength--;
        }
    }
}

3, deformation title

There are two sorted array a1 and a2, a1 at the end of memory has enough free time receiving a2. Implement a function, all numbers a2 a2 insert, and all of the numbers are sorted.

///需要排序的两个数组和其各自的长度
void Sort(int arr1[], int arr2[], int length1, int length2)
{
    if (arr1 == nullptr || arr2 == nullptr || length2<=0)
    {
        return;
    }
    
    //总长度
    int length = length1 + length2;

    //从后向前依次比较
    while (length1 >= 0 && length > length1)
    {
        if (arr1[length1-1] >= arr2[length2-1])
        {
            arr1[length - 1] = arr1[length1 - 1];
            length--;
            length1--;
        }
        else {
            arr1[length - 1] = arr2[length2 - 1];
            length--;
            length2--;
        }
    }
}