在剛接觸到這個題目的時候,看完題目完全的不自所云,所以在網上借鑑了很多解法,才明白了大意:
思路1:將第一個結點算出然後循環剩餘結點
/**
* Definition for singly-linked list.
* function ListNode(val) {
* this.val = val;
* this.next = null;
* }
*/
/**
* @param {ListNode} l1
* @param {ListNode} l2
* @return {ListNode}
*/
var addTwoNumbers = function(l1, l2) {
var sum=parseInt(l1.val)+parseInt(l2.val);
var tmp=sum>=10?1:0;
var pionF=l1.next;
var poinS=l2.next;
var data=new ListNode(sum%10);
var nextList=data;
while(pionF || poinS || tmp){
if(pionF || poinS){
sum=parseInt(pionF.val)+parseInt(poinS.val)+tmp;
pionF=pionF.next;
poinS=poinS.next;
}else{
sum=tmp;
}
tmp=sum>=10?1:0;
var info=new ListNode(sum%10);
nextList.next=info;
nextList=nextList.next;
}
return data;
};
結果:當然是沒有通過,因爲沒有考率到兩個輸入的長度可能不一樣的問題
思路2:從開始就循環,使用三目運算判斷賦值
/**
* Definition for singly-linked list.
* function ListNode(val) {
* this.val = val;
* this.next = null;
* }
*/
/**
* @param {ListNode} l1
* @param {ListNode} l2
* @return {ListNode}
*/
var addTwoNumbers = function(l1, l2) {
var tmp=0;
var list=0;
var data=new ListNode(0);
var poin=data;
while(l1 || l2 || tmp){
list++;
var val1=l1?(l1.val!=null?l1.val:0):0;
var val2=l2?(l2.val!=null?l2.val:0):0;
var sum=val1+val2+tmp;
tmp=sum>=10?1:0;
var info=new ListNode(sum%10);
poin.next=info;
poin=poin.next;
l1=l1?(l1.next!=null?l1.next:null):null;
l2=l2?(l2.next!=null?l2.next:null):null;
}
return data.next;
};
結果:是通過的但是這個時間卻是很長。。。
思路3:因爲思路2的結果上LeetCode提示了!=與!==的提示,所以做以下的修改
/**
* Definition for singly-linked list.
* function ListNode(val) {
* this.val = val;
* this.next = null;
* }
*/
/**
* @param {ListNode} l1
* @param {ListNode} l2
* @return {ListNode}
*/
var addTwoNumbers = function(l1, l2) {
var tmp=0;
var list=0;
var data=new ListNode(0);
var poin=data;
while(l1 || l2 || tmp){
list++;
var val1=l1?(l1.val!==null?l1.val:0):0;
var val2=l2?(l2.val!==null?l2.val:0):0;
var sum=val1+val2+tmp;
tmp=sum>=10?1:0;
var info=new ListNode(sum%10);
poin.next=info;
poin=poin.next;
l1=l1?(l1.next!==null?l1.next:null):null;
l2=l2?(l2.next!==null?l2.next:null):null;
}
return data.next;
};
結果:出乎意料,!=和!==的執行效率區別有那麼大麼?黑人臉????在JavaScript中==的實現算法要複雜與===,==時候雙方類型的不確定所以包含了轉型操作
思路4:好吧,承認我拿的是LeetCode答案前面最快的(185ms),但是實際的速度則沒有那麼快,可能這是很久以前的答案吧,數據有了變動,所以。。。。
/**
* Definition for singly-linked list.
* function ListNode(val) {
* this.val = val;
* this.next = null;
* }
*/
/**
* @param {ListNode} l1
* @param {ListNode} l2
* @return {ListNode}
*/
var addTwoNumbers = function(l1, l2) {
carry = 0
var p1 = l1
var p2 = l2
var head = new ListNode(-1)
var p = head
while (p1 || p2)
{
var v1 = 0
if (p1)
{
v1 = p1.val
p1 = p1.next
}
var v2 = 0
if (p2)
{
v2 = p2.val
p2 = p2.next
}
var s = v1+v2+carry
if (s>9)
{
carry = 1
s -= 10
}
else
{
carry = 0
}
p.next = new ListNode(s)
p = p.next
}
if (carry)
{
p.next = new ListNode(carry)
}
return head.next
};
結果:並沒有標註出來的那麼快
思路5:遞歸,這也是前面的答案(189ms)
/**
* Definition for singly-linked list.
* function ListNode(val) {
* this.val = val;
* this.next = null;
* }
*/
/**
* @param {ListNode} l1
* @param {ListNode} l2
* @return {ListNode}
*/
var addTwoNumbers = function(l1, l2) {
const res = test(l1, l2, [], false);
return res;
};
var test = function (l1, l2, res, overflow = false) {
res = res || [];
l1 = l1 || { val: 0, next: null };
l2 = l2 || { val: 0, next: null };
const val = l1.val + l2.val + Number(overflow);
if (val >= 10) {
res.push(val % 10);
overflow = true;
} else {
res.push(val);
overflow = false;
}
if (l1.next || l2.next || overflow) {
test(l1.next, l2.next, res, overflow);
}
return res;
};
結果:也是並沒有那麼快