1758. Minimum Changes To Make Alternating Binary String

题目：

You are given a string s consisting only of the characters '0' and '1'. In one operation, you can change any '0' to '1' or vice versa.

The string is called alternating if no two adjacent characters are equal. For example, the string "010" is alternating, while the string "0100" is not.

Return the minimum number of operations needed to make s alternating.

Example 1:

Input: s = "0100"
Output: 1
Explanation: If you change the last character to '1', s will be "0101", which is alternating.

Example 2:

Input: s = "10"
Output: 0
Explanation: s is already alternating.

Example 3:

Input: s = "1111"
Output: 2
Explanation: You need two operations to reach "0101" or "1010".

Constraints:

1 <= s.length <= 10^4
s[i] is either '0' or '1'.

思路：

按照题意，alternating字符串无非是0101或者1010，那么我们就创造一下这两种字符串，和原字符串等长，然后与原字符串对比得出不同的位置数再返回最小值即可。同时这里可以边创建字符串边和原字符串比较，因此two pass即可。

代码：

class Solution {
public:
int minOperations(string s) {
int n=s.size();
int c1=0, c2=0;
string s1="";
for(int i=0;i<n;i++)
{
if(i%2==0)
s1+='0';
else
s1+='1';
if(s1[i]!=s[i])
c1++;
}
string s2="";
for(int i=0;i<n;i++)
{
if(i%2==0)
s2+='1';
else
s2+='0';
if(s2[i]!=s[i])
c2++;
}
return min(c1,c2);
}
};

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