方法一:
//清除字符串中的空格
// 1.需要计算空格个数
// 2.将字符串平移
// 3.由于平移之后字符串后面会出现重复,将字符串中的重复的第一个赋值为'\0';在主程序中打印时只打印'\0'之前的
public static char[] remove(String str, char ch) {
char[] arr = str.toCharArray();
int count = 0;
for (int i = 0; i < arr.length-1; i++) {
if (arr[i] == ch)
count++;
}
for (int i = 0; i < arr.length; i++) {
if (arr[i] == ch) {
for (int j = i; j < arr.length - 1; j++)
arr[j] = arr[j + 1];
}
}
arr[arr.length-count]='\0';
return arr;
}
public static void main(String[] args){
String str="I want to eat";
System.out.println(str);//打印原字符串
char[]arr=remove(str,' ');
int i=0;
while (arr[i]!='\0'){
System.out.print((arr[i]));//打印清除空格之后的的字符串
i++;
}
}
由于上面程序存在数组空间浪费,即只在主程序中打印arr 的length-count(空格个数)的长度;
方法二:对上面程序改进,在定义函数中,根据空格个数对原数组长度进行缩容
public static char[] remove(String str, char ch) {
char[] arr = str.toCharArray();
int count = 0;
for (int i = 0; i < arr.length-1; i++) {
if (arr[i] == ch)
count++;
}
for (int i = 0; i < arr.length; i++) {
if (arr[i] == ch) {
for (int j = i; j < arr.length - 1; j++)
arr[j] = arr[j + 1];
}
}
char[] brr=Arrays.copyOf(arr,arr.length-count);
return brr;
}
public static void main(String[] args){
String str="I want to eat";
System.out.println(str);
char[]arr=remove(str,' ');
System.out.print(Arrays.toString(arr));
}