1.爬楼梯：斐波那契数列的算法
class Solution {public: /** * @param n: An integer * @return: An integer */ int climbStairs(int n) { // write your code here int fib1=1; int fib2=2; if(n==0) { return 1; } if(n==1) { return 1; } if(n==2) { return 2; } if(n>1) { for(int i=2;i 
2.排列序号：对原数组中的每一个位置,找到其后有多少个数比它小,由于字典排序是从小到大排列的，所以只需计算出第i位（从1开始）在它后面比它小的数进行(A.size()-k-1)!，最后相加即可。
class Solution {public: /** * @param A an integer array * @return a long integer */ long long permutationIndex(vector & A) { // Write your code here int i,j; long n=1; for(i=0;i A[j]) { long m=1; for(int k=i;A.size()-k-1>0;k++) { m=m*(A.size()-k-1); } n=n+m; } } } return n; }}; 
class Solution {public: /** * @param A and B: sorted integer array A and B. * @return: A new sorted integer array */ vector mergeSortedArray(vector &A, vector &B) { // write your code here vector C(A.size()+B.size(), 0); int i,j; for(i=0;i j) { C[m]=C[m-1]; m--; } C[j]=k; p++; } } } return C; }}; 
/** * Definition of ListNode * * class ListNode { * public: * int val; * ListNode *next; * * ListNode(int val) { * this->val = val; * this->next = NULL; * } * } */class Solution {public: /** * @param head: The first node of linked list. * @return: The new head of reversed linked list. */ ListNode *reverse(ListNode *head) { // write your code here ListNode *prev=NULL; while(head!=NULL) { ListNode *t=head->next; head->next=prev; prev=head; head=t; } return prev; }}; 
class Solution {public: /** * @param nums: a vector of integers * @return: an integer */ int maxProduct(vector & nums) { // write your code here int Max = nums[0]; int Min = nums[0]; int allMax = nums[0]; for(int i=1;i //自己写的第一种但复杂度过高
/*int i,j,k,m; k=nums[0]; for(i=0;i
class Solution {public: /** * @param s a string, encoded message * @return an integer, the number of ways decoding */ int numDecodings(string& s) { // Write your code here int fib1=1,fib2=1; if (s.empty()||s.front()=='0') { return 0; } for(int i=1;i 
/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */class Solution {public: /** * @param headA: the first list * @param headB: the second list * @return: a ListNode */ ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) { // write your code here stack A,B; ListNode *x=NULL; if(headA==NULL||headB==NULL) { return NULL; } else { for(;headA!=NULL;headA=headA->next) { A.push(headA); } for(;headB!=NULL;headB=headB->next) { B.push(headB); } for(;A.top()!=NULL&&B.top()!=NULL;) { if(A.top()==B.top()) { x=A.top(); A.pop(); B.pop(); } else { break; } } return x; } }}; 
class Solution {public: /** * Get all distinct N-Queen solutions * @param n: The number of queens * @return: All distinct solutions * For example, A string '...Q' shows a queen on forth position */ vector > solveNQueens(int n) { // write your code here vector > ret; search(0,n,ret); return ret; } int c[20]; void search(int r,int &n, vector > &ret) { if(r == n) { vector v; for(int i=0; i 
class Solution {public: /** * @param nums: A list of integers. * @return: A list of permutations. */ vector > permute(vector nums) { // write your code here vector > result; int size = nums.size(); permute(nums, 0, size, result); return result; } void permute(vector &nums, int begin, int end, vector > &result) { if(begin == end) { result.push_back(nums); } else { for(int i=begin; i 
class Solution {public: /** * @param s : A string * @return : A string */ string reverseWords(string s) { // write your code here string word, line; int size=s.size(); int i=size-1, j, wordbegin=size-1, wordend=size-1; if(s.empty()) { return s; } for(;i>=0;) { for(;s[i]==' '&&i>=0;i--) { wordbegin--; wordend--; } for(;s[i]!=' '&&i>=0;i--) { wordbegin--; } word.resize(0); for(j=wordbegin+1; j<=wordend; j++) { word.append(1,s[j]); } if(wordbegin==0) { line=line+word; } else { line=line+word+" "; } wordend =wordbegin; } return line; } }; 