sword_offer 面试题5：替换空格

/**
* 替换空格
* 题目：请实现一个函数，把字符串中的每个空格替换成“%20”，例如
* 输入“we are happy”，则输出“we%20are%20happy”.
*/
public static String replaceBlank(String str){
if (str == null || str.length()<=0) {
return "";
}
StringBuilder sb = new StringBuilder();
for (int i = 0; i < str.length(); i++) {
if (str.charAt(i) != ' ') {
sb.append(str.charAt(i));
}else {
sb.append("%20");
}
}
return sb.toString();
}
    /**
     * 相關題目：有两个排序的数组A1和A2，请实现一个函数，把A1和A2合并成一个排序数组。
     * 解题思路：1、新建一个辅助数组，长度为A1+A2，把A1和A2中的所有数字先全部放入到辅助数组，然后再进行排序
     * @param arr
     * @param brr
     * @return
     */
public static int[] mergeStr1(int[] arr,int[] brr){
int arrLength = arr.length;
int brrLength = brr.length;
int[] crr = new int[arrLength+brrLength];
if (arr==null || arr.length<=0) {
return brr;
}else if (brr==null || brr.length<=0) {
return arr;
}else if ((arr==null || arr.length<=0) || (brr==null || brr.length<=0)) {
return crr;
}
for (int i = 0; i < crr.length; i++) {
if (i<arrLength) {
crr[i] = arr[i];
}else {
crr[i] = brr[i-arrLength];
}
}
for (int i = 0; i < crr.length; i++) {
for (int j = i+1; j < crr.length; j++) {
if (crr[i]>crr[j]) {
int tmp = crr[i];
crr[i] = crr[j];
crr[j] = tmp;
}
}
}
return crr;
}
/**
* 解法思路2：直接先比较两个数组的值，然后塞入辅助数组
* @param arr
* @param brr
* @return
*/
public static int[] mergeStr2(int[] arr,int[] brr){
int arrLength = arr.length;
int brrLength = brr.length;
int[] crr = new int[arrLength+brrLength];
if (arr==null || arr.length<=0) {
return brr;
}else if (brr==null || brr.length<=0) {
return arr;
}else if ((arr==null || arr.length<=0) || (brr==null || brr.length<=0)) {
return crr;
}
int i=0,k=0,j=0;
while(i<arrLength && j <brrLength){
if (arr[i] <= brr[j]) {
crr[k++] = arr[i++];
}else {
crr[k++] = brr[j++];
}

}

              //这两步操作是把比较完之后，arr或者brr剩余的数依次添加进辅助数组

while(i<arrLength){
crr[k++] = arr[i++];
}
while(j<brrLength){
crr[k++] = brr[j++];
}
return crr;
}