Letter Combinations of a Phone Number
这题iterate方法的基本思路就是连续做+1，+1的含义就是移动同一数字对应字母下一个index处。算法是两层循环，外层是每一个输出，内层是一个数的每一位。难点是循环的invariant：外层是carry!=1，因为如果内层循环最高位仍有进位说明已经过了最后一个对应string。内层从低到高遍历每一位，但如果carry变为0，要break中止。
错误点
class Solution(object):def letterCombinations(self, digits):""":type digits: str:rtype: List[str]"""def getOneRes(indices):res = []for i in range(len(indices)):res.append(dmap[ord(digits[i])-ord('0')][indices[i]])return ''.join(res)if not digits: return []dmap = ["", "", "abc","def","ghi","jkl","mno","pqrs","tuv","wxyz"]indices = [0]*len(digits)carry = 0res = []while not carry:res.append(getOneRes(indices))for i in range(len(digits)-1, -1, -1):indices[i] += 1if indices[i]==len(dmap[ord(digits[i])-ord('0')]):indices[i]=0carry=1else:carry=0breakreturn res
