题目
Given two binary strings, return their sum (also a binary string).
For example, a = “11”, b = “1”, Return “100”.
解答
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| public class Solution {
public String addBinary(String a, String b) {
int acur = a.length() - 1;
int bcur = b.length() - 1;
StringBuffer sb = new StringBuffer();
boolean flag = false;
while(acur >=0 || bcur >= 0){
int anum = (acur >= 0 ? a.charAt(acur) - '0' : 0);
int bnum = (bcur >= 0 ? b.charAt(bcur) - '0' : 0);
int res = anum + bnum;
if (flag == true) {
res++;
flag = false;
}
if (res > 1) {
res = res - 2;
flag = true;
}
sb.append(res);
acur--;
bcur--;
}
if(flag == true){
sb.append('1');
}
return sb.reverse().toString();
}
}
|
注解
没啥好说的，flag标志位标识是否进位，字符串从后往前依次取字符相加，并通过StringBuider追加，最后翻转字符串，返回结果。
反思
CharAt()函数的底层实现
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| public char charAt(int index) {
if ((index < 0) || (index >= count)) {
throw new StringIndexOutOfBoundsException(index);
}
return value[index + offset];
}
|
StringBuilder与StringBuffer
StringBuilder vs StringBuffer: StringBuffer is synchronized, which means it is thread-safe but slower than StringBuilder.
