306. Additive Number

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本文介绍了一种通过三重循环来判断一个仅包含数字的字符串是否为加成数的方法。加成数是指该字符串的数字可以形成至少三个数的序列，且从第三个数开始每个数都是前两个数之和。文章提供了详细的逻辑分析及通过LeetCode验证的代码实现。

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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.

思路:就是三重循环判断

代码如下(已通过leetcode)

public class Solution {
public boolean isAdditiveNumber(String num) {

for (int i = 0; i < num.length() - 2; i++) {
long a;
if(num.substring(0,i+1).charAt(0)=='0'&&num.substring(0,i+1).length()!=1) break;
else a = Long.valueOf(num.substring(0, i + 1));

for (int j = i + 1; j < num.length() - 1; j++) {
long b;
if(num.substring(i+1,j+1).charAt(0)=='0'&&num.substring(i+1,j+1).length()!=1) break;
else b = Long.valueOf(num.substring(i + 1, j + 1));

for (int k = j + 1; k < num.length(); k++) {
long c;
if(num.substring(j+1,k+1).charAt(0)=='0'&&num.substring(j+1,k+1).length()!=1) break;
else c = Long.valueOf(num.substring(j + 1, k + 1));
//System.out.println("a: " + a + " " + "b: " + b + " " + "c: " + c);
if (a + b == c) {
boolean flag = true;
int temp = k + 1;
if(k+1>=num.length()) return true;
int len=0;
while (temp+len <=num.length()) {
c=a+b;
//System.out.println("a: " + a + " " + "b: " + b + " " + "c: " + c);

a = b;
b = c;

long count = a + b;
len = (count + "").length();
if (a + b != Long.valueOf(num.substring(temp, temp + len))) {
flag = false;
break;
} else {
temp = temp + len;
}

}
if (flag)
return true;
} else {
a=Long.valueOf(num.substring(0, i + 1));
b=Long.valueOf(num.substring(i + 1, j + 1));
if(a+b<c) break;
}
}
}

}
return false;
}
}