資訊專(zhuān)欄INFORMATION COLUMN
摘要：感想剛開(kāi)始看到這道題，覺(jué)得很簡(jiǎn)單，跟歸并的過(guò)程比較像，算法復(fù)雜度。可以半小時(shí)之內(nèi)完全可以搞定。可是隨著一次次提交出問(wèn)題，發(fā)現(xiàn)似乎沒(méi)有我想的那么簡(jiǎn)單問(wèn)題其實(shí)就出現(xiàn)在各種沒(méi)有處理好，劍指多次提到的寫(xiě)代碼要注意健壯性深刻體會(huì)到了。
You are given two linked lists representing two non-negative numbers. The digits are stored in reverse order and each of their nodes contain 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
感想： 剛開(kāi)始看到這道題，覺(jué)得很簡(jiǎn)單，跟歸并merge的過(guò)程比較像，算法復(fù)雜度O（n）。可以半小時(shí)之內(nèi)完全可以搞定。可是隨著一次次提交出問(wèn)題，發(fā)現(xiàn)似乎沒(méi)有我想的那么簡(jiǎn)單...問(wèn)題其實(shí)就出現(xiàn)在各種corner cases沒(méi)有處理好，which is also 劍指offer多次提到的寫(xiě)代碼要注意健壯性....深刻體會(huì)到了。
思路： 處理好進(jìn)位，要自己手動(dòng)寫(xiě)各種情況的測(cè)試用例，確保程序的健壯性...
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) {
* val = x;
* next = null;
* }
* }
*/
public class Solution {
private int getSum(int x, int y, int digit){
int value = x+y+digit;
if(value>=10)
return value-10;
else
return value;
}
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
ListNode p = l1;
ListNode q = l2;
// new head node for the resultlist
ListNode newHead = new ListNode(getSum(p.val,q.val,0));
ListNode pHead = newHead;
int digit;//處理進(jìn)位
if(p.val+q.val-10 >= 0){
digit = 1;
}
else
digit = 0;
p = p.next;
q = q.next;
while(p!=null || q != null){
// 兩個(gè)鏈表長(zhǎng)度相等時(shí)候的處理
if(p!=null &&q!=null){
ListNode temp = new ListNode(getSum(p.val, q.val,digit));
pHead.next = temp;
pHead = pHead.next;
if(p.val+q.val+digit-10 >= 0){
digit = 1;
}
else
digit = 0;
p = p.next;
q = q.next;
}
// 兩個(gè)鏈表長(zhǎng)度不等時(shí)候的處理
else if(p==null){
ListNode temp = new ListNode(getSum(0, q.val,digit));
pHead.next = temp;
pHead = pHead.next;
if(0+q.val-10+digit >= 0){
digit = 1;
}
else{
digit = 0;
}
q = q.next;
}
// 兩個(gè)鏈表長(zhǎng)度不等時(shí)候的處理
else if(q==null){
ListNode temp = new ListNode(getSum(p.val, 0,digit));
pHead.next = temp;
pHead = pHead.next;
if(0+p.val-10+digit >= 0){
digit = 1;
}
else{
digit= 0;
}
p = p.next;
}
}
// 鏈表1： 5 鏈表2： 5 這個(gè)時(shí)候的處理
if(digit ==1){
ListNode temp = new ListNode(getSum(0, 0,digit));
pHead.next = temp;
}
return newHead;
}
}
更優(yōu)美一點(diǎn)的代碼
javapublic class Solution {
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
int carry =0;
ListNode newHead = new ListNode(0);
ListNode p1 = l1, p2 = l2, p3=newHead;
while(p1 != null || p2 != null){
if(p1 != null){
carry += p1.val;
p1 = p1.next;
}
if(p2 != null){
carry += p2.val;
p2 = p2.next;
}
p3.next = new ListNode(carry%10);
p3 = p3.next;
carry /= 10;
}
if(carry==1)
p3.next=new ListNode(1);
return newHead.next;
}
}
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