LeetCode 67th title: binary summation (simple)

LeetCode 67th title: binary summation (simple)

• Title: Given two binary string, and return to their (expressed in binary). And a non-empty input string contains only numbers 0 and 1.
• An idea: the two strings turn into an integer, are added, then the obtained and converted to binary. But there is a long string exceeds the limit.

class Solution {
    public String addBinary(String a, String b) {
        int aa = Integer.parseInt(a);
        int bb = Integer.parseInt(b);
        int sum = aa+bb;
        String ans = "";
        if(sum == 0) return "0";
        while(sum>0){
            int k=sum%10;
            if(k < 2){
                sum=sum/10;
                ans+=k;
            }else{
                sum=sum/10+1;
                k=k-2;
                ans+=k;
            }
        }
        return new StringBuilder(ans).reverse().toString();
    }
}

• Ideas II: From the start of the last string judge.

class Solution {
    public String addBinary(String a, String b) {
        int lenA = a.length();
        int lenB = b.length();
        int aa = lenA-1;
        int bb = lenB-1;
        String ans = "";
        String next = "0";
        while(aa>=0 && bb>=0){
            if(a.charAt(aa)=='1' && b.charAt(bb)=='1'){
                ans+=next;
                next="1";
            }else if(a.charAt(aa)=='1' || b.charAt(bb)=='1'){
                if(next=="1"){
                    ans+="0";
                    next="1";
                }else{
                    ans+="1";
                    next="0";
                }
            }else{
                ans+=next;
                next="0";
            }
            aa--;
            bb--;
        }


        if(aa>=0){
            while(aa>=0){
                if(next=="1"){
                    if(a.charAt(aa)=='0'){
                        ans+="1";
                        next="0";
                    }else{
                        ans+="0";
                        next="1";
                    }
                }else{
                    ans+=a.charAt(aa);
                }
                aa--;                
            }
        }else if(bb>=0){
            while(bb>=0){
                if(next=="1"){
                    if(b.charAt(bb)=='0'){
                        ans+="1";
                        next="0";
                    }else{
                        ans+="0";
                        next="1";
                    }
                }else{
                    ans+=b.charAt(bb);
                }
                bb--;
            }
        }
        if(next=="1") ans+="1";

        return  new StringBuilder(ans).reverse().toString();
    }
}

• Thinking three: problem solution in the short string padded with zeros, I saved half of the code.

class Solution {
    public String addBinary(String a, String b) {
        StringBuilder ans = new StringBuilder();
        int ca = 0;
        for(int i = a.length() - 1, j = b.length() - 1;i >= 0 || j >= 0; i--, j--) {
            int sum = ca;
            sum += i >= 0 ? a.charAt(i) - '0' : 0;
            sum += j >= 0 ? b.charAt(j) - '0' : 0;
            ans.append(sum % 2);
            ca = sum / 2;
        }
        ans.append(ca == 1 ? ca : "");
        return ans.reverse().toString();
    }
}

作者：guanpengchn
链接：https://leetcode-cn.com/problems/add-binary/solution/hua-jie-suan-fa-67-er-jin-zhi-qiu-he-by-guanpengch/
来源：力扣（LeetCode）
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。