leetcode415

改进前的代码：
public class leetcode415 {
    private StringBuilder res = new StringBuilder();
    public String addStrings(String num1, String num2) {
        int length1 = num1.length();
        int length2 = num2.length();
        int count = 1;
        boolean isAddOne = false;
        while (length1 - count >=0 && length2 - count >= 0) {
            int temp = num1.charAt(length1-count) - '0' + num2.charAt(length2-count)-'0';
            if (isAddOne) {
                temp++;
            }
            if (temp < 10) {
                res.append(temp);
                isAddOne = false;
            } else {
                res.append(temp%10);
                isAddOne = true;
            }
            count++;
        }
        if (length1 == length2 && isAddOne) {
            res.append(1);
        } else if (length1 > length2) {
            processMul(num1,length1,count,isAddOne);
        } else {
            processMul(num2,length2,count,isAddOne);
        }

        return res.reverse().toString();
    }

    private void processMul(String s, int length,int count,boolean isAddOne) {
        while (length - count >=0) {
            int temp = s.charAt(length-count) - '0';
            if (isAddOne) {
                temp++;
            }
            if (temp < 10) {
                res.append(temp);
                isAddOne = false;
            } else {
                res.append(temp%10);
                isAddOne = true;
            }
            count++;
        }
        if (isAddOne) {
            res.append(1);
        }
    }
}

改进点：
1.较短的那个字符串可以在前面补0，不用重新执行那个循环！

第一次改进后：
public class leetcode415 {
    public String addStrings(String num1, String num2) {
        int length1 = num1.length();
        int length2 = num2.length();
        int count = 1;
        StringBuilder res = new StringBuilder();
        boolean isAddOne = false;
        while (length1 - count >=0 || length2 - count >= 0) {
            int number1 = length1 - count >= 0?num1.charAt(length1 - count)-'0':0;
            int number2 = length2 - count >= 0?num2.charAt(length2 - count)-'0':0;
            int temp = number1 + number2;
            if (isAddOne) {
                temp++;
            }
            if (temp < 10) {
                res.append(temp);
                isAddOne = false;
            } else {
                res.append(temp%10);
                isAddOne = true;
            }
            count++;
        }
        if (isAddOne) {
            res.append(1);
        }

        return res.reverse().toString();
    }
}

改进点：
1.length1 - count 和 length2 - count 没必要，可以直接用两个变量来代替。
2.不用区分temp是否小于10，res.append(temp%10)都成立。
3.isAddOne 的赋值可以用三元运算符。

第二次改进后：
public class leetcode415 {
    public String addStrings(String num1, String num2) {
        int p1 = num1.length() - 1;
        int p2 = num2.length() - 1;
        StringBuilder res = new StringBuilder();
        boolean isAddOne = false;
        while (p1 >=0 || p2 >= 0) {
            int number1 = p1 >= 0?num1.charAt(p1)-'0':0;
            int number2 = p2 >= 0?num2.charAt(p2)-'0':0;
            int temp = number1 + number2;
            if (isAddOne) {
                temp++;
            }
            res.append(temp%10);
            isAddOne = temp>=10;
            p1--;
            p2--;
        }
        if (isAddOne) {
            res.append(1);
        }

        return res.reverse().toString();
    }
}