[leetcode #2] -- Add two Numbers

題目

Add Two Numbers
You are given two linked lists representing two non-negative numbers. The digits are stored in reverse order and each of their nodes contains a single digit. Add the two numbers and return it as a linked list.

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8

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分析

難度中等.
這個題目注意一下是reverse order 意思就是 2 -> 4 -> 3分別是個位,十位,百位
而且l1 & l2的長度不必定是相同的.
解決這道題目的主要是理解兩個數相加最大的是進1,不可能進2或者更大的.定義一個carry來記錄不進位仍是1就行了.

解法

var addTwoNumbers = function(l1, l2) {
    let carry = 0;
    let first = new ListNode();
    let l3 = first;
    
    while(l1 || l2 || carry) {  
    // carry 是爲了檢驗如l1 {5} l2{5}這種狀況,就是
    //  最高位要進位的那種
    //  假如不添加,那麼 最高位要是進位的話不會被考慮進來.
    //  若是不明白能夠把while裏面的carry去掉 而後測試一下
        let sum = carry;
        if (l1 !== null) {
            sum += l1.val;
            l1 = l1.next;
        }
        if (l2 !== null) {
            sum += l2.val;
            l2 = l2.next;
        }
    
        if (sum > 9) {
            carry = 1;
            sum = sum-10;
        }
        else {
            carry = 0;
        }
        l3.next = new ListNode(sum);
        l3 = l3.next;
    }
    
    return first.next;
};