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89 lines
3.1 KiB
Markdown
Executable File
89 lines
3.1 KiB
Markdown
Executable File
# [306. Additive Number](https://leetcode.com/problems/additive-number/)
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## 题目
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Additive number is a string whose digits can form additive sequence.
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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.
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Given a string containing only digits `'0'-'9'`, write a function to determine if it's an additive number.
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**Note**: Numbers in the additive sequence **cannot** have leading zeros, so sequence `1, 2, 03` or `1, 02, 3` is invalid.
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**Example 1**:
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Input: "112358"
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Output: true
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Explanation: The digits can form an additive sequence: 1, 1, 2, 3, 5, 8.
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1 + 1 = 2, 1 + 2 = 3, 2 + 3 = 5, 3 + 5 = 8
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**Example 2**:
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Input: "199100199"
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Output: true
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Explanation: The additive sequence is: 1, 99, 100, 199.
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1 + 99 = 100, 99 + 100 = 199
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**Follow up**:How would you handle overflow for very large input integers?
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## 题目大意
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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。
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## 解题思路
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- 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。
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- 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。
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- 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。
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## 代码
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```go
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package leetcode
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import (
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"strconv"
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"strings"
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)
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// This function controls various combinations as starting points
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func isAdditiveNumber(num string) bool {
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if len(num) < 3 {
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return false
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}
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for firstEnd := 0; firstEnd < len(num)/2; firstEnd++ {
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if num[0] == '0' && firstEnd > 0 {
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break
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}
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first, _ := strconv.Atoi(num[:firstEnd+1])
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for secondEnd := firstEnd + 1; max(firstEnd, secondEnd-firstEnd) <= len(num)-secondEnd; secondEnd++ {
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if num[firstEnd+1] == '0' && secondEnd-firstEnd > 1 {
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break
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}
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second, _ := strconv.Atoi(num[firstEnd+1 : secondEnd+1])
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if recursiveCheck(num, first, second, secondEnd+1) {
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return true
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}
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}
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}
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return false
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}
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//Propagate for rest of the string
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func recursiveCheck(num string, x1 int, x2 int, left int) bool {
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if left == len(num) {
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return true
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}
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if strings.HasPrefix(num[left:], strconv.Itoa(x1+x2)) {
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return recursiveCheck(num, x2, x1+x2, left+len(strconv.Itoa(x1+x2)))
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}
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return false
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}
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``` |