题目链接
注意点
- 考虑开头数字有进位的情况
- 如何将string和int之间转化
解法
解法一:carry表示当前是否有进位,从尾部开始逐位相加。时间复杂度O(n)
class Solution {
public:
string addBinary(string a, string b) {
string ret = "";
int i = a.length()-1,j = b.length()-1;
int carry = 0;
while(i >= 0 && j >= 0)
{
int temp = (a[i]+b[j]+carry)-96;
if(temp>= 2) carry = 1;
else carry = 0;
ret = to_string(temp%2)+ret;
i--;
j--;
}
while(i >= 0)
{
int temp = (a[i]+carry)-48;
if(temp >= 2) carry = 1;
else carry = 0;
ret = to_string(temp%2)+ret;
i--;
}
while(j >= 0)
{
int temp = (b[j]+carry)-48;
if(temp >= 2) carry = 1;
else carry = 0;
ret = to_string(temp%2)+ret;
j--;
}
if(carry == 1) ret = "1"+ret;
return ret;
}
};
解法二:解法一中to_string效率太低了,仔细思考会发现其实真正可能出现的数字只有4个:0、1、2、3
,所以就自己写个函数实现。时间复杂度O(n)
class Solution {
public:
char toChar(int num)
{
if(num == 0) return '0';
else if(num == 1) return '1';
else if(num == 2) return '0';
else if(num == 3) return '1';
return '0';
}
string addBinary(string a, string b) {
string ret = "";
int i = a.length()-1,j = b.length()-1;
int carry = 0;
while(i >= 0 && j >= 0)
{
int temp = (a[i]+b[j]+carry)-96;
if(temp>= 2) carry = 1;
else carry = 0;
ret.insert(ret.begin(),toChar(temp));
i--;
j--;
}
while(i >= 0)
{
int temp = (a[i]+carry)-48;
if(temp >= 2) carry = 1;
else carry = 0;
ret.insert(ret.begin(),toChar(temp));
i--;
}
while(j >= 0)
{
int temp = (b[j]+carry)-48;
if(temp >= 2) carry = 1;
else carry = 0;
ret.insert(ret.begin(),toChar(temp));
j--;
}
if(carry == 1) ret = "1"+ret;
return ret;
}
};
小结
- 模拟的思想,模拟手工加法的过程。