题目:

二进制求和

给定两个二进制字符串,返回他们的和(用二进制表示)。

样例

a = 11

b = 1

返回 100

解题:

和求两个链表的和很类似

考虑进位,考虑最后一项的进位

0+0 = 0 不需要进位

0+1 = 1 不需要进位

1+1 =0  进位 1

同时注意

低位进1,高位时1+1的情况,直接加就是3了,这个需要进位1 ,原位的结果也是1的情况

Java程序:

public class Solution {
/**
* @param a a number
* @param b a number
* @return the result
*/
public String addBinary(String a, String b) {
// Write your code here
String result = "";
int aLen = a.length() - 1;
int bLen = b.length() - 1;
int sum = 0;
while(aLen>=0 || bLen>=0){
if(aLen>=0){
sum +=Integer.parseInt(a.substring(aLen,aLen+1));
aLen--;
}
if(bLen>=0){
sum +=Integer.parseInt(b.substring(bLen,bLen+1));
bLen--;
}
if(sum==2){
result = "0" + result;
sum=1;
}else if(sum==0 || sum==1) {
result = sum +"" + result;
sum = 0;
}else if(sum==3){
result = "1" + result;
sum = 1;
}
}
if(sum==1)
result = "1" + result;
return result; }
}

Python程序:
总耗时: 10156 ms

class Solution:
# @param {string} a a number
# @param {string} b a number
# @return {string} the result
def addBinary(self, a, b):
# Write your code here
result =''
aLen = len(a) - 1
bLen = len(b) - 1
sum = 0
while aLen>=0 or bLen>=0:
if aLen>=0:
sum += int(a[aLen])
aLen-=1
if bLen>=0:
sum+= int(b[bLen])
bLen-=1
if sum==3:
result = '' + result
sum=1
elif sum==2:
result = '' + result
sum=1
else:
result = str(sum) + result
sum = 0
if sum==1:
result = '' + result
return result

总耗时: 523 ms

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