LeetCode #445: Add Two Numbers II
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Problem
You are given two non-empty linked lists representing two non-negative integers. The most significant digit comes first and each of their nodes contain a single digit. Add the two numbers and return it as a linked list.
You may assume the two numbers do not contain any leading zero, except the number 0 itself.
Follow up:
What if you cannot modify the input lists? In other words, reversing the lists is not allowed.
Example:
Input: (7 -> 2 -> 4 -> 3) + (5 -> 6 -> 4)Output: 7 -> 8 -> 0 -> 7
Solution
题意
给定两个数字(以链表形式给出),求它们的和。
比如:7243+564 = 7807
题目会给出两个链表,分别为7 -> 2 -> 4 -> 3 与 5 -> 6 -> 4,要求得到两个数的和的链表形式。
思路
由于链表的性质,实现两个数的加法,一种方法就是从最后一位开始加,从个位往上,依次进位。
可以利用栈来实现这一目的。
将两个链表分别压入两个栈,s1和s2中,用carry表示进位,则每一位的结果就是s1.top() + s2.top() + carry(其中某个栈为空的情况下,则可以视作sx.top() = 0;在代码中则需要条件控制)
如果该位的结果大于等于10,则进位carry=1,留给下一位的计算;否则,carry=0。
计算完该位后两个栈分别pop一次,pop时要注意判断当前栈是否已空,防止程序崩溃。
Code
#include<stack>class Solution {public: ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) { //两个栈分别存放l1和l2 stack<int> s1, s2; ListNode * l1_s1 = l1; ListNode * l2_s2 = l2; while(l1_s1 != NULL){ s1.push(l1_s1->val); l1_s1 = l1_s1->next; } while(l2_s2 != NULL){ s2.push(l2_s2->val); l2_s2 = l2_s2->next; } //由于给定的两个链表非空,所以个位一定存在,先构造个位的node ListNode * node = new ListNode(s1.top() + s2.top()); //进位 int carry = 0; //如果>=10,则进位 if (node->val >= 10){ carry = node->val / 10; node->val %= 10; } s1.pop(); s2.pop(); while(!(s1.empty() && s2.empty())){ //由于ListNode没有提供默认构造函数,所以用0暂时代替,不会影响之后的构造 previousNode = new ListNode(0); if (s1.empty()){ previousNode->val = s2.top() + carry; s2.pop(); } else if (s2.empty()){ previousNode->val = s1.top() + carry; s1.pop(); } else{ previousNode->val = s1.top() + s2.top() + carry; s1.pop(); s2.pop(); } if (previousNode->val >= 10){ carry = previousNode->val / 10; previousNode->val %= 10; } else carry = 0; //构造链表 previousNode->next = node; node = previousNode; } //例如5+5,pop一次后,两个栈都空,但carry不为0,就需要再构造一个高位 if (s1.empty() && s2.empty() && carry != 0){ previousNode = new ListNode(carry); previousNode->next = node; node = previousNode; } return node; }};
Summary
- 虽然难度是Medium,但是代码实现并不是特别复杂,但是对于条件的判断需要细心一点,否则容易造成Runtime Error
- 想到借助栈来实现从链表尾部开始的加法也需要过程
- 本周写了三道题。
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