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website/content/ChapterFour/0306.Additive-Number.md

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+# [306. Additive Number](https://leetcode.com/problems/additive-number/)
+
+
+## 题目
+
+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.
+
+Given a string containing only digits `'0'-'9'`, write a function to determine if it's an additive number.
+
+**Note**: Numbers in the additive sequence **cannot** have leading zeros, so sequence `1, 2, 03` or `1, 02, 3` is invalid.
+
+**Example 1**:
+
+    Input: "112358"
+    Output: true 
+    Explanation: 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
+
+**Example 2**:
+
+    Input: "199100199"
+    Output: true 
+    Explanation: The additive sequence is: 1, 99, 100, 199. 
+                 1 + 99 = 100, 99 + 100 = 199
+
+**Follow up**:How would you handle overflow for very large input integers?
+
+
+## 题目大意
+
+累加数是一个字符串，组成它的数字可以形成累加序列。一个有效的累加序列必须至少包含 3 个数。除了最开始的两个数以外，字符串中的其他数都等于它之前两个数相加的和。给定一个只包含数字 '0'-'9' 的字符串，编写一个算法来判断给定输入是否是累加数。说明: 累加序列里的数不会以 0 开头，所以不会出现 1, 2, 03 或者 1, 02, 3 的情况。
+
+
+## 解题思路
+
+- 在给出的字符串中判断该字符串是否为斐波那契数列形式的字符串。
+- 由于每次判断需要累加 2 个数字，所以在 DFS 遍历的过程中需要维护 2 个数的边界，`firstEnd` 和 `secondEnd`，两个数加起来的和数的起始位置是 `secondEnd + 1`。每次在移动 `firstEnd` 和 `secondEnd` 的时候，需要判断 `strings.HasPrefix(num[secondEnd + 1:], strconv.Itoa(x1 + x2))`，即后面的字符串中是否以和为开头。
+- 如果第一个数字起始数字出现了 0 ，或者第二个数字起始数字出现了 0，都算非法异常情况，都应该直接返回 false。
+
+
+
+## 代码
+
+```go
+
+package leetcode
+
+import (
+	"strconv"
+	"strings"
+)
+
+// This function controls various combinations as starting points
+func isAdditiveNumber(num string) bool {
+	if len(num) < 3 {
+		return false
+	}
+	for firstEnd := 0; firstEnd < len(num)/2; firstEnd++ {
+		if num[0] == '0' && firstEnd > 0 {
+			break
+		}
+		first, _ := strconv.Atoi(num[:firstEnd+1])
+		for secondEnd := firstEnd + 1; max(firstEnd, secondEnd-firstEnd) <= len(num)-secondEnd; secondEnd++ {
+			if num[firstEnd+1] == '0' && secondEnd-firstEnd > 1 {
+				break
+			}
+			second, _ := strconv.Atoi(num[firstEnd+1 : secondEnd+1])
+			if recursiveCheck(num, first, second, secondEnd+1) {
+				return true
+			}
+		}
+	}
+	return false
+}
+
+//Propagate for rest of the string
+func recursiveCheck(num string, x1 int, x2 int, left int) bool {
+	if left == len(num) {
+		return true
+	}
+	if strings.HasPrefix(num[left:], strconv.Itoa(x1+x2)) {
+		return recursiveCheck(num, x2, x1+x2, left+len(strconv.Itoa(x1+x2)))
+	}
+	return false
+}
+
+```