牛客 NC1 大数加法(模拟)

将两个字符串逆置
然后维护s为最长串进行遍历相加 注意是否有进位

public class Solution {
    
    
    int carry = 0;
    public void add(StringBuilder sb ,char x ,char y){
    
    
        int sum = x -'0' + y - '0' + carry;//相加
        Integer tmp = sum%10;
        String c = tmp.toString(); 
        sb.append(c);//追加字符串
        carry = sum/10;//维护进位
    }
    public String solve (String s, String t) {
    
    
        if(s.length() < t.length()){
    
    
            String tmp = s;
            s = t;
            t = tmp;
        }//维护s为最长字符串
        s = new StringBuffer(s).reverse().toString();//反转字符串
        t = new StringBuffer(t).reverse().toString();//反转字符串
        StringBuilder sb = new StringBuilder();
        for(int i = 0;i < t.length();i++){
    
    
            add(sb,s.charAt(i),t.charAt(i));
        }
        for(int i = t.length();i < s.length();i++){
    
    
            add(sb,s.charAt(i),'0');
        }
        if(carry == 1) sb.append('1');//如果有进位追加1
        return sb.reverse().toString();
    }
}