替換空格
題目:請實現一個函數,將一個字符串中的空格替換成“%20”。例如,當字符串爲We Are Happy.則經過替換之後的字符串爲We%20Are%20Happy。
輸入:”We Are Happy.”
輸出:”We%20Are%20Happy.”
思路:1、先遍歷一遍字符串,找到原字符數組的長度和空格的個數
2、然後對於每個空格爲字符數組長度+2,因爲原空格佔1個字符,而加的%20是佔3個字符
3、然後設置2個指針,一個指向新的位置,一個指向老的位置,依次進行復制,遇到空格時,老位置只減1,而新位置依次加入‘0’、‘2’、‘%’即可
4、最終只需要遍歷2次該數組,即時間複雜度爲O(n)
/*
實現:替換空格,例子是"we are happy.",空格替換爲'%20'
*/
#include<iostream>
#include<string>
using namespace std;
//length表示字符型數組最大的長度
void ReplaceBlank(char c[],int length){
if (c == NULL){
return;
}
int blankNumber = 0;
int originalLength = 0;
int i = 0;
while (c[i] != '\0'){
originalLength++;
if (c[i] == ' '){
blankNumber++;
}
i++;
}
if (originalLength <= 0){
return;
}
int newLength = originalLength + blankNumber * 2;
if (newLength > length){
return;
}
int newIndex = newLength;
int originIndex = originalLength;
while (originIndex >= 0){
if (c[originIndex] != ' '){
c[newIndex--] = c[originIndex--];
}
else{
c[newIndex--] = '0';
c[newIndex--] = '2';
c[newIndex--] = '%';
originIndex--;
}
}
cout << c;
}
int main(){
char c[] = "we are happy.";
ReplaceBlank(c,100);
system("pause");
return 0;
}