2. Add Two Numbers
You are given two non-empty linked lists representing two non-negative integers. 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.
You may assume the two numbers do not contain any leading zero, except the number 0 itself.
Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
Explanation: 342 + 465 = 807.
C++
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
ListNode* p=new ListNode(0); //answer
ListNode* top=p;
ListNode* q=NULL; //temp
int flag =0;
int tmp;
while(l1&&l2){
tmp = l1->val+l2->val+flag;
flag = tmp/10;
q = new ListNode(tmp%10);
p->next = q;
p = p->next;
l1 = l1->next;
l2 = l2->next;
}
while(l1){
tmp = l1->val+flag;
flag = tmp/10;
q = new ListNode(tmp%10);
p->next = q;
p = p->next;
l1 = l1->next;
}
while(l2){
tmp = l2->val+flag;
flag = tmp/10;
q = new ListNode(tmp%10);
p->next = q;
p = p->next;
l2 = l2->next;
}
if(flag>0){
q = new ListNode(flag);
p->next = q;
}
return top->next;
}
};
4. Median of Two Sorted Arrays
There are two sorted arrays nums1 and nums2 of size m and n respectively.
Find the median of the two sorted arrays. The overall run time complexity should be O(log (m+n)).
Example 1:
nums1 = [1, 3]
nums2 = [2]
The median is 2.0
Example 2:
nums1 = [1, 2]
nums2 = [3, 4]
The median is (2 + 3)/2 = 2.5
class Solution {
private:
static int find_kth(vector<int>::const_iterator A, int m,
vector<int>::const_iterator B, int n, int k) {
//always assume that m is equal or smaller than n
if (m > n) return find_kth(B, n, A, m, k);
if (m == 0) return *(B + k - 1);
if (k == 1) return min(*A, *B);
//divide k into two parts
int ia = min(k / 2, m), ib = k - ia;
if (*(A + ia - 1) < *(B + ib - 1))
return find_kth(A + ia, m - ia, B, n, k - ia);
else if (*(A + ia - 1) > *(B + ib - 1))
return find_kth(A, m, B + ib, n - ib, k - ib);
else
return A[ia - 1];
}
public:
double findMedianSortedArrays(const vector<int>& A, const vector<int>& B) {
int m = A.size();
int n = B.size();
int total = m + n;
if (total & 0x1) //判斷奇數(shù)
return find_kth(A.begin(), m, B.begin(), n, total / 2 + 1);
else
return (find_kth(A.begin(), m, B.begin(), n, total / 2) +
find_kth(A.begin(), m, B.begin(), n, total / 2 + 1)) / 2.0;
}
};
三種情況
? A[k/2-1] == B[k/2-1] 找到答案了
? A[k/2-1] > B[k/2-1] 第K項不在B前k/2-1內(nèi)
? A[k/2-1] < B[k/2-1] 第K項不在A前k/2-1內(nèi)遞歸停止條件
? m==0晨逝， 從B里邊直接算第K個即可
? k ==1， 找到兩個元素中更小的
? A[k/2-1] == B[k/2-1] 找到答案了
思路之巧妙
