Java小练习-替换空格

请实现一个函数，将一个字符串中的每个空格替换成“%20”。例如，当字符串为We Are Happy.则经过替换之后的字符串为We%20Are%20Happy.

(1) 第一种解法是重新开辟一个数组,实现相应的替换

    public static int getCount(char[] arr, char src) {
        int count = 0;
        for (int i = 0; i < arr.length; i++) {
            if (arr[i] == src) {
                count++;
            }
        }
        return count;
    }

    public static char[] replace(char[] arr, char src, int num, char aim) {
        if (arr == null) {
            return null;
        }
        int begin = 0;
        int end = arr.length - 1;
        int srcCount = getCount(arr, src);                      // 获取需要替换目标的个数
        char[] brr = new char[arr.length + srcCount * (num - 1)];
        int i = 0;
        int j = brr.length - 1;
        while (begin <= end) {                                     //begin end 为原数组的两头下标   i  j 为新数组的两头下标
            if (arr[begin] != src && begin <= end) {
                brr[i] = arr[begin];
                begin++;                    //原数组下标加一 新数组下标加一
                i++;
            } else {
                begin++;                            //替换时,原数组只加一次
                for (int k = 0; k < num; k++) {      //新数组在循环中加
                    brr[i] = aim;               
                    i++;
                }
            }
            if (arr[end] != src && end > begin) {
                if (arr[end] != src) {
                    brr[j] = arr[end];
                    end--;
                    j--;
                } else {
                    end--;
                    for (int k = num; k > 0; k--) {
                        brr[j] = aim;
                        j--;
                    }
                }
            }
        }
        return brr;
    }
    public static void main(String[] args) {
        long startTime = System.currentTimeMillis();
//do something
        String str = "i am a student!";
        char[] arr = str.toCharArray();
        arr = replace(arr, ' ', 2, '#');
        System.out.println(Arrays.toString(arr));
        long endTime = System.currentTimeMillis();
        System.out.println("程序运行时间：" + (endTime - startTime) + "ms");
    }

(2) 在原数组的基础上进行扩容,通过系统的扩容函数进行扩容,这种方法在,内存消耗与上一种方法差别不大,无论是Arrays.CopyOf()方法还是System.arrayCopy()在底层对于一维数组来说都是深拷贝,这种方法的难点是将数组移位.

public class replaceBlankMove {
    public static int getCount(char[] arr, int src) {
        int count = 0;
        for (int i = 0; i < arr.length; i++) {
            if (arr[i] == src) {
                count++;
            }
        }
        return count;
    }
    public static char[] replace(char[] arr, char src, int num, char aim) {
        if (arr == null) {
            return null;
        }
        int arrCount = arr.length - 1;             //保存arr最开始的末尾值
        int srcCount = getCount(arr, src);
        arr = Arrays.copyOf(arr, arr.length + srcCount * (num - 1));

        for (int i = 0; i < arr.length; i++) {
            if (arr[i] == src) {

                for (int j = arrCount; j > i; j--) {                        //先移位将i后面的元素移动num-1位
                    arr[j + num - 1] = arr[j];
                }
                for (int k = i; k <= i + num - 1; k++) {
                    arr[k] = aim;
                }
                arrCount += num - 1;                                    //在此处将添加移动后的尾巴
            }
        }
        return arr;
    }

    public static void main(String[] args) {
        String str = "i am a student!";
        char[] arr = str.toCharArray();
        System.out.println(replace(arr, ' ', 2, '#'));
    }

代码一次运行成功的感觉是最爽的.