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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.
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?
思路
利用backtracking 和recursive即可解决，这里要注意的是：大数用BigDecimal来解决。
实现代码如下：
private List<List<BigDecimal>> allRes = new ArrayList<List<BigDecimal>>();public boolean isAdditiveNumber(String num) {help(num,new ArrayList<BigDecimal>(),0); return allRes.size()>0;}private void help(String num, ArrayList<BigDecimal> res, int pos) {if(pos==num.length()){//int size = res.size();if(size>=3){if(res.get(size-1).subtract(res.get(size-2)).equals(res.get(size-3))){//这个判断可以不要allRes.add(res);}}}for(int i=pos;i<num.length();i++){if(i!=pos&&num.charAt(pos)=='0'){//多位数的开头不能为零break;}BigDecimal val = new BigDecimal(num.substring(pos, i+1));int size = res.size(); if(size<2){ res.add(val);help(num, new ArrayList(res),i+1);res.remove(res.get(res.size()-1));//将最后一个添加进来的元素删除}else{if(val.subtract(res.get(size-1)).equals(res.get(size-2))){res.add(val);help(num, new ArrayList(res),i+1);res.remove(res.get(res.size()-1));//将最后一个添加进来的元素删除}}}}这篇关于《leetcode》：Additive Number的文章就介绍到这儿，希望我们推荐的文章对编程师们有所帮助！
