力扣题:高精度运算-1.4

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力扣题-1.4

[力扣刷题攻略] Re：从零开始的力扣刷题生活

力扣题1：306. 累加数

解题思想：首先先通过secondStart和secondEnd可以确定num1 = num[0:secondStart],num2 = num[secondStart:secondEnd]，然后遍历secondStart和secondEnd进行第二个数字的提取，然后通过check函数进行判断是否合法。

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class Solution(object):def isAdditiveNumber(self, num):""":type num: str:rtype: bool"""n = len(num)## 遍历secondStart和secondEnd即可，因为num1 = num[0:secondStart],num2 = num[secondStart:secondEnd]## secondStart从1开始，secondEnd从secondStart+1开始for secondStart in range(1,n-1):## 判断特殊条件num1=0if num[0] == '0' and secondStart!=1:breakfor secondEnd in range(secondStart,n-1):## 判断特殊条件num2=0if num[secondStart] == '0' and secondStart != secondEnd:breaknum1 = num[0:secondStart]num2 = num[secondStart:secondEnd+1]num3 = numif self.check(num1,num2,num3):return Truereturn Falsedef add(self,num1,num2):## 进行字符串的加法操作，从低位开始相加i, j = len(num1) - 1, len(num2) - 1add = 0ans = list()while i >= 0 or j >= 0 or add != 0:x = int(num1[i]) if i >= 0 else 0y = int(num2[j]) if j >= 0 else 0result = x + y + addans.append(str(result % 10))add = result // 10i -= 1j -= 1## 最后需要进行翻转return "".join(ans[::-1])def check(self,num1,num2,num3):n = len(num3)num3 = num3[len(num1)+len(num2):]flag = 1while num3 != "" and flag == 1:temp = self.add(num1,num2)temp_len = len(temp)curr_num = num3[:temp_len]if curr_num == temp:num1 = num2num2 = tempnum3 = num3[temp_len:]else:flag = 0if flag == 0:return Falsereturn True

class Solution {
public:bool isAdditiveNumber(string num) {int n = num.length();for (int secondStart = 1; secondStart < n - 1; ++secondStart) {if (num[0] == '0' && secondStart != 1) {break;}for (int secondEnd = secondStart; secondEnd < n - 1; ++secondEnd) {if (num[secondStart] == '0' && secondStart != secondEnd) {break;}string num1 = num.substr(0, secondStart);string num2 = num.substr(secondStart, secondEnd - secondStart + 1);string num3 = num;if (check(num1, num2, num3)) {return true;}}}return false;}string add(string num1, string num2) {int i = num1.length() - 1;int j = num2.length() - 1;int add = 0;string result = "";while (i >= 0 || j >= 0 || add != 0) {int x = (i >= 0) ? (num1[i] - '0') : 0;int y = (j >= 0) ? (num2[j] - '0') : 0;int sum = x + y + add;result = char(sum % 10 + '0') + result;add = sum / 10;i -= 1;j -= 1;}return result;}bool check(string num1, string num2, string num3) {int n = num3.length();num3 = num3.substr(num1.length() + num2.length());int flag = 1;while (!num3.empty() && flag == 1) {string temp = add(num1, num2);int tempLen = temp.length();string currNum = num3.substr(0, tempLen);if (currNum == temp) {num1 = num2;num2 = temp;num3 = num3.substr(tempLen);} else {flag = 0;}}return (flag == 1);}
};