leetCode第二題答案，兩個鏈表存儲形式的數據加和，最終返回一個鏈表（java版分析）

題目詳情

我的開始代碼解析

這個代碼是簡單實現版，沒有追求代碼的精簡。

package com.ke.leetCodes;


class ListNode {
    int val;
     ListNode next;
     ListNode(int x) { val = x; }
  }
 
//Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
//Output: 7 -> 0 -> 8
//Explanation: 342 + 465 = 807.

class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
    	if(l1==null) return l2;
    	if(l2==null) return l1;
    	ListNode t1 = l1;
    	ListNode t2 = l2;
    	ListNode result = null;
    	ListNode head = null;
    	int carry = 0;
    	int data = 0;
    	while(t1!=null&&t2!=null){
    		data = (t1.val+t2.val+carry)%10;  
    		carry = (t1.val+t2.val+carry)/10; //進位
    		ListNode current = new ListNode(data);
    		
    		if(result == null) {
    			result = current;
    			head = result;
    		}else{
    			result.next = current;
    			result = result.next;
    		}
    		t1 = t1.next;
    		t2 = t2.next;
    	}
    	while(t1!=null){
    		data = (t1.val+carry)%10;  
    		carry = (t1.val+carry)/10; //進位
    		ListNode current = new ListNode(data);
    		if(result == null) {
    			result = current;
    			head = result;
    		}else{
    			result.next = current;
    			result = result.next;
    		}
    		t1 = t1.next;
    	}
    	while(t2!=null){
    		data = (t2.val+carry)%10;  
    		carry = (t2.val+carry)/10; //進位
    		ListNode current = new ListNode(data);
    		if(result == null) {
    			result = current;
    			head = result;
    		}else{
    			result.next = current;
    			result = result.next;
    		}
    		t2 = t2.next;
    	}
    	if(carry!=0){
    		ListNode current = new ListNode(carry);
    		result.next = current;
			result = result.next;
    	}
    	
    	
    	return head;
        
    }
    
  //Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
  //Output: 7 -> 0 -> 8
  //Explanation: 342 + 465 = 807.
    public static void main(String[] args) {
//		int [] a1 = new int[]{2,4,3};
//		int [] a2 = new int[]{5,6,4};
		int [] a1 = new int[]{1};
		int [] a2 = new int[]{9,9};
		ListNode l1 = null;
		ListNode l2 = null;
		ListNode h1 = null;
		ListNode h2 = null;
		for(int i = 0;i<a1.length;i++){
			if(l1 == null){
				h1 = new ListNode(a1[i]);
				l1= h1;
			}else{
				h1.next =  new ListNode(a1[i]);
				h1 = h1.next;
			}
		}
		for(int i = 0;i<a2.length;i++){
			if(l2 == null){
				h2 = new ListNode(a2[i]);
				l2 = h2;
			}else{
				h2.next =  new ListNode(a2[i]);
				h2 = h2.next;
			}
		}
		
		print(l1);
		print(l2);
		Solution obj = new Solution();
		Solution.print(obj.addTwoNumbers(l1, l2));
    	
	}
    public static void print(ListNode l){
    	ListNode tl= l;
    	while(tl!=null){
    		System.out.print(tl.val+" ");
    		tl=tl.next;
    	}
    	System.out.println();
    }
}

精選答案

如果不看別的大神寫的代碼，永遠都不會提高：
答案。

public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
    ListNode dummyHead = new ListNode(0);
    ListNode p = l1, q = l2, curr = dummyHead;
    int carry = 0;
    while (p != null || q != null) {
        int x = (p != null) ? p.val : 0;
        int y = (q != null) ? q.val : 0;
        int sum = carry + x + y;
        carry = sum / 10;
        curr.next = new ListNode(sum % 10);
        curr = curr.next;
        if (p != null) p = p.next;
        if (q != null) q = q.next;
    }
    if (carry > 0) {
        curr.next = new ListNode(carry);
    }
    return dummyHead.next;
}

可以從參考答案中借鑑的點

1. 將表頭直接新建了個值爲0的節點，這樣保證頭結點是可以直接next，而不拋空指針異常。
2. 循環語句的精簡：既然都是一遍大循環，爲什麼循環不能用在一個裏面呢？
3. 最後的進位，仍然要進行判斷，如果有進位，要再新加一個節點，進行鏈接。