leetcode306

class Solution:
    def backTrack(self,num,l,idx,n):
        if idx >= n:
            if len(l) >= 3:
                return True
            else:
                return False
        for i in range(idx,n):
            s = num[idx:i+1]
            leng = len(s)
            cur = int(s)
            s_cur = str(cur)
            if s_cur != s:
                continue#有前导0
            if len(l) < 2:
                l.append(cur)
                res = self.backTrack(num,l,idx+leng,n)
                if res:
                    return True
                else:
                    if len(l) != 0:
                        l.pop(-1)
            else:
                pre = l[-1]
                prepre = l[-2]
                if pre + prepre == cur:
                    l.append(cur)
                    res = self.backTrack(num,l,idx+leng,n)
                    if res:
                        return True
                    else:
                        if len(l) != 0:
                            l.pop(-1)
        return False


    def isAdditiveNumber(self, num: str) -> bool:
        n = len(num)
        return self.backTrack(num,[],0,n)

算法思路：回溯法。