摘要：題目解答不越界長度的當可以走到后面沒有和了的時候，說明這個滿足條件直接可以知道這個是不是存在于中越界長度的越界長的度

題目：
Additive number is a string whose digits can form additive sequence.

A valid additive sequence should contain at least three numbers. Except for the first two numbers, each subsequent number in the sequence must be the sum of the preceding two.

For example:
"112358" is an additive number because the digits can form an additive sequence: 1, 1, 2, 3, 5, 8.

1 + 1 = 2, 1 + 2 = 3, 2 + 3 = 5, 3 + 5 = 8
"199100199" is also an additive number, the additive sequence is: 1, 99, 100, 199.
1 + 99 = 100, 99 + 100 = 199
Note: Numbers in the additive sequence cannot have leading zeros, so sequence 1, 2, 03 or 1, 02, 3 is invalid.

Given a string containing only digits "0"-"9", write a function to determine if it"s an additive number.

Follow up:
How would you handle overflow for very large input integers?

解答：
不越界長度的Recursive:

public boolean isValid(Long x1, Long x2, String num, int start) {
    //當可以走到后面沒有和了的時候，說明這個string滿足條件
    if (start == num.length()) return true;
    x2 = x1 + x2;
    x1 = x2 - x1;
    String sum = x2.toString();
    //string.startsWith()直接可以知道這個sum是不是存在于num中
    return num.startsWith(sum, start) && isValid(x1, x2, num, start + sum.length());
}

public boolean isAdditiveNumber(String num) {
    int len = num.length();
    for (int i = 1; i <= len / 2; i++) {
        if (num.charAt(0) == "0" && i > 1) return false;
        Long x1 = Long.parseLong(num.substring(0, i));
        for (int j = 1; Math.max(i, j) <= len - i - j; j++) {
            if (num.charAt(i) == "0" && j > 1) break;
            Long x2 = Long.parseLong(num.substring(i, i + j));
            if (isValid(x1, x2, num, i + j)) return true;
        }
    }
    return false;
}

越界長度的recursive:

public boolean isValid(BigInteger x1, BigInteger x2, String num, int start) {
    if (start == num.length()) return true;
    x2 = x2.add(x1);
    x1 = x2.subtract(x1);
    String sum = x2.toString();
    return num.startsWith(sum, start) && isValid(x1, x2, num, start + sum.length());
}

public boolean isAdditiveNumber(String num) {
    int len = num.length();
    for (int i = 1; i <= len / 2; i++) {
        if (num.charAt(0) == "0" && i > 1) return false;
        BigInteger x1 = new BigInteger(num.substring(0, i));
        for (int j = 1; Math.max(i, j) <= len - i - j; j++) {
            if (num.charAt(i) == "0" && j > 1) break;
            BigInteger x2 = new BigInteger(num.substring(i, i + j));
            if (isValid(x1, x2, num, i + j)) return true;
        }
    }
    return false;
}

越界長的度iterator: (tail recursion is easy to transfer)

public boolean isValid(int i, int j, String num) {
    if (num.charAt(0) == "0" && i > 1) return false;
    if (num.charAt(i) == "0" && j > 1) return false;
    BigInteger x1 = new BigInteger(num.substring(0, i));
    BigInteger x2 = new BigInteger(num.substring(i, i + j));
    String sum;
    for (int start = i + j; start < num.length(); start += sum.length()) {
        x2 = x2.add(x1);
        x1 = x2.subtract(x1);
        sum = x2.toString();
        if (!num.startsWith(sum, start)) return false;
    }
    return true;
}

public boolean isAdditiveNumber(String num) {
    int len = num.length();
    for (int i = 1; i <= len / 2; i++) {
        for (int j = 1; Math.max(i, j) <= len - i - j; j++) {
            if (isValid(i, j, num)) {
                return true;
            }
        }
    }
    return false;
}