1、leetcode #28 implement strStr()
Topic address: https: //leetcode.com/problems/implement-strstr/
Analysis: Looking to achieve a string function, namely to achieve similar substr () function
If the string is not found, returns -1, find the index returns a match pattern string of the first character
If a matching input string is empty, returns 0
Methods: KMP matching algorithm
Note: When the main string when empty output haystack -1 instead of 0
int strStr(string haystack, string needle) { if(needle == "") return 0; //generate next[] int next[needle.length()] = {0}; next[0] = 0; int temp = 0; for(int i = 1; i < needle.length(); i++){ temp = next[i-1]; while(needle[i] != needle[temp] && temp > 0){ temp = next[temp-1]; } if(needle[i] == needle[temp]) next[i] = temp + 1; else next[i] = 0; } //KMP search int hp = 0,np = 0; for(;hp < haystack.length() ; hp++){ while(haystack[hp] != needle[np] && np) np = next[np - 1]; if(haystack[hp] == needle[np]) np++; if(np == needle.length()){ return hp - np + 1; } } //didn't find return -1; }
2. leetcode #38 count_and_say
Topic address: https: //leetcode.com/problems/count-and-say/
Recursion
string countAndSay(int n){ if(n == 1) return "1"; else{ string str = countAndSay(n-1); string result = ""; char c; int counter = 0 , i = 0; while(i < str.length()){ c = str[i]; counter ++; while(str[++i] == c){ counter++; } result += to_string(counter) + c; counter = 0; } return result; } }
3. FIG topics are as follows
Analysis: First, a need to break the string into a word, the isolated word onto the stack
工具：#include<sstream> #include<stack>
#include <iostream> #include <sstream> #include <stack> using namespace std; void reverse_word_by_word(string str); int main(){ string str; getline(cin,str); reverse_word_by_word(str); } void reverse_word_by_word(string str){ stringstream ss; string temp_word; stack<string>result; ss.str(str); //split words and push then in a stack while((ss >> temp_word) && !ss.fail()){ result.push(temp_word); } //pop these words cout << result.top(); result.pop(); while(!result.empty()){ cout << " " << result.top() ; result.pop(); } cout << endl; }
4. FIG topics are as follows
Analysis: analog adder string of numbers, taking into account a possible long string of numbers, such as 10000 "1", c ++ biggest variable is long long int, if simply convert a string to a number, there may be overflow problem, so this method is adopted discarded analog adder carry method ... ----
Note: This algorithm only considers the integer arithmetic, real number not consider
#include <iostream> #include <sstream> #include <stack> #include <cstring> using namespace std; /* ## analysis #1 length not match, eg. "12345" + "456" = "12801" #2 all 0's, "000" + "0" = "0" #3 one of them is empty, eg. "" + "123" = "123" #4 the final overflow, eg. "9999" + "1" = "100000" #5 invalid input ## note # the Hex conting method is ot implemented */ int sum_of_two_strings(string str1, string str2, int counting_method = 10){ if(str1 == "" && str2 == ""){ cout << "Invalid input!" << endl; return -1; } //####chose counting method char S ,E; switch(counting_method){ case 10: S='0';E='9';break; case 8: S='0';E='7';break; case 2: S='0';E='1';break; default:cout << "Not supported counting method!"<<endl;return -1; } stack<char>result; bool overflow = 0; int temp_add; int i = str1.length()-1 , j = str2.length()-1; //right alignment, calculate till the short one consumed for(; i >= 0 && j >= 0; i-- , j--){ if(str1[i] < S || str1[i] > E || str2[j] < S || str2[j] > E){ printf("%s\n", "Invalid input!"); return -1; }else{ if(overflow){ temp_add = (str1[i] - S) + (str2[j] - S) + 1; if(temp_add > counting_method-1){ overflow = 1; result.push(temp_add % counting_method ); }else{ overflow = 0; result.push(temp_add); } }else{ temp_add = (str1[i] - S) + (str2[j] - S); if(temp_add > counting_method-1){ overflow = 1; result.push(temp_add % counting_method ); }else{ overflow = 0; result.push(temp_add); } } } } //if str1 bigger then str2, calculate the rest of it while(i>= 0){ if(!(S < str1[i] <= E)){ printf("%s\n", "Invalid input!"); return -1; }else{ if (overflow) { int temp_add = (str1[i] - S) + 1; if (temp_add > counting_method-1) { overflow = 1; result.push(temp_add % counting_method); }else{ overflow = 0; result.push(temp_add); } }else{ result.push((str1[i])-S); } i--; } } //if str2 bigger then str1, calculate the rest of it while(j >= 0){ if(!(S < str2[j] <= E)){ printf("%s\n", "Invalid input!"); return -1; }else{ if (overflow) { int temp_add = (str2[j] - S) + 1; if (temp_add > counting_method-1) { overflow = 1; result.push(temp_add % counting_method ); }else{ overflow = 0; result.push(temp_add); } }else{ result.push((str2[j])- S); } j--; } } //consider the final overflow //eg.99999 + 1 = 1000000(D), the overflow exits throughout the calculation period if(overflow){ result.push(1); } //get the naive answer, it may include 000018 which the prefix of 0's we don't want char result_str[result.size()]; int index = result.size()-1; while(!result.empty()){ result_str[index--] = (char)(result.top() + S); result.pop(); } index = sizeof(result_str) / sizeof(char); //consume the prefix of 0's while((--index >= 0) && result_str[index] == S){;} //consider 000 + 0 , we can not consume all the 0's but stay ont 0 if(index == -1) cout << S; else{ //after correct consumed 0's, put the final answer while(index >= 0){ cout << result_str[index--]; } } cout << endl; return 1; } int main(){ string str1, str2; int counting_method, times = 10; for(int i = 1; i <= times ; i++){ cout << "########## test:" << i << " ###########" << endl; cout << "Chose counting method(2,8,10,16):"; cin >> counting_method; getchar(); cout << "str1:"; getline(cin, str1); cout << "str2:"; getline(cin, str2); sum_of_two_strings(str1, str2,counting_method); cout << endl; } }
5 leetcode #75 sort colors
Topic link: https: //leetcode.com/problems/sort-colors/
Methods: Three-way quick sort
void sortColors(vector<int>& nums) { int len = nums.size(); int one_index = 0, two_index = len; int pointer = 0; while(pointer < two_index){ if(nums[pointer] == 0){ swap(nums[pointer++], nums[one_index++]); }else if(nums[pointer] == 2){ swap(nums[pointer],nums[--two_index]); }else{pointer++;} } }
6. leetcode #167 two sum(2)
Topic address: https: //leetcode.com/problems/two-sum-ii-input-array-is-sorted/
vector<int> twoSum(vector<int>& numbers, int target) { vector<int> result; int i = 0; int j = numbers.size() - 1; //int mid = j - ( j - i ) / 2; while(i < j){ if(numbers[i] + numbers[j] == target){ result.push_back(i+1); result.push_back(j+1); return result; }else if(numbers[i] + numbers[j] < target){ i ++ ; }else{ j --; } } return result; }
7. leetcode #283 move_zeroes
Topic address: https: //leetcode.com/problems/move-zeroes/
void moveZeroes(vector<int>& nums) { int pointer = 0; for(int i = 0; i < nums.size(); i++){ if(nums[i] != 0){ nums[pointer++] = nums[i]; } } while(pointer < nums.size()){ nums[pointer++] = 0; } }
8. leetcode #51 N-Queens
Topic link: https: //leetcode.com/problems/n-queens/
Algorithm Analysis: A Case Study of 4 Queen
void N_Queens(std::vector<std::vector<string>> &result, std::vector<string> &NQueen, std::vector<int> &flag, int row , int &n ){ if(row == n){ result.push_back(NQueen); return; }else{ for(int col = 0; col != n; col ++){ if(flag[col] && flag[n + col + row] && flag[n + (2 * n -1) + (n - 1 + col - row)]){ flag[col] = flag[n + col + row] = flag[4 * n - 2 + col - row] = 0; NQueen[row][col] = 'Q'; N_Queens(result, NQueen, flag, row + 1, n); NQueen[row][col] = '.'; flag[col] = flag[n + col + row] = flag[4 * n - 2 + col - row] = 1; } } } } std::vector<std::vector<string>> solveNQueens(int n){ std::vector<string> NQueen(n, string(n, '.')); std::vector<int> flag(5 * n - 2, 1); std::vector<std::vector<string>> result; N_Queens(result, NQueen, flag , 0, n); return result; }
9 leetcode #2 add two numbers
Topic link: https: //leetcode.com/problems/add-two-numbers/
Analysis of Algorithms:
Method1
if l1 and l2 are in the same size, no attention
if l1 is longer than l2, attention the overflow, eg. [9,9] + [1] = [0,0,1] not [0,10].
if l2 is longer than l1,put the result of l2 to the tail of result list
Method2
recursion
ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) { //nullptr judgement if(l1 == nullptr){ return l2?l2:NULL; } if(l2 == nullptr) return l1?l1:NULL; //teh head and tail of result number list ListNode* head = l1 ,*tail = l1; while(l1 && l2 ){ int temp_add = l1->val + l2->val; if(temp_add > 9){ //avoid using >= 10, that needs more time to execute temp_add %= 10; if(l1->next) l1->next->val++; else l1->next = new ListNode(1); } l1->val = temp_add; tail = l1; l1 = l1->next; l2 = l2->next; } //l1 is the longer one, if current val greater than 9, execute this block while(l1 && l1->val > 9){ if(l1->next) l1->next->val++; else l1->next = new ListNode(1); l1->val = l1->val % 10; l1 = l1->next; } // l2 is the longer one if(l2){ tail->next = l2; } return head; }
ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) { if(l1 == nullptr){ return l2?l2:NULL; } if(l2 == nullptr) return l1?l1:NULL; int sum = l1->val + l2->val; if(sum > 9){ sum = sum % 10; if( l1->next ){ l1->next->val++; }else{ l1->next = new ListNode(1); } if(l2->next == nullptr) l2->next = new ListNode(0); } ListNode* cur_node = new ListNode(sum); cur_node->next = addTwoNumbers(l1->next, l2->next); return cur_node; }
10 leetcode # 4 median of two sorted array
Topic link: https: //leetcode.com/problems/median-of-two-sorted-arrays/
Algorithm Analysis: The merge sort and heap sort. Merge sort using the next procedure in FIG. Heap sort using a large stack top and the top of the heap of small, intermediate values recorded median.
//using merge sort double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) { vector<int> nums; nums.resize(nums1.size() + nums2.size()); merge(nums1.begin(),nums1.end(),nums2.begin(),nums2.end(),nums.begin() ); int half_size = nums.size() / 2; if(nums.size() % 2){ return nums[half_size] * 1.0; }else{ return (nums[half_size - 1] + nums[half_size]) / 2.0; } } //using heap sort double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) { if(nums1.empty()){ if(nums2.size() % 2 == 0) return (nums2[nums2.size() / 2 -1] + nums2[nums2.size() / 2]) / 2.0; else return nums2[nums2.size() / 2] * 1.0; } if(nums2.empty()){ if(nums1.size() % 2 == 0) return (nums1[nums1.size() / 2 -1] + nums1[nums1.size() / 2]) / 2.0; else return nums1[nums1.size() / 2] * 1.0; } multiset<int> left; //min heap multiset<int> right; //max heap vector<int> temp; //make sure nums1 is the longest vector, so that we can create two heap faster if(nums1.size() < nums2.size()){ temp = nums1; nums1 = nums2; nums2 = temp; } int index = 0; int median; int nums1_half = nums1.size() / 2; if(nums1.size() % 2){ index = 0; median = nums1[nums1_half]; while(index < nums1_half) left.insert(nums1[index++]); while(++index < nums1.size()) right.insert(nums1[index]); }else{ index = 0; median = nums1[nums1_half - 1]; while(index < nums1_half - 1) left.insert(nums1[index++]); while(++index < nums1.size()) right.insert(nums1[index]); } index = 0; while(index < nums2.size()){ if(right.size() > left.size() && right.size() - left.size() > 1){ left.insert(median); median = *right.begin(); right.erase(right.begin()); } if(left.size() > right.size() && left.size() - right.size() > 1){ right.insert(median); multiset<int>::iterator iter = --left.end(); median = *iter; left.erase(iter); } if(nums2[index] >= median){ right.insert(nums2[index]); }else{ left.insert(nums2[index]); } index ++ ; } if(right.size() > left.size() && right.size() - left.size() > 1){ left.insert(median); median = *right.begin(); right.erase(right.begin()); } if(left.size() > right.size() && left.size() - right.size() > 1){ right.insert(median); multiset<int>::iterator iter = --left.end(); median = *iter; left.erase(iter); } if(right.size() == left.size()){ return median; }else if(right.size() > left.size()){ return (median + *(right.begin())) / 2.0; }else{ return (median + *(--left.end())) /2.0; } }
11 leetcode # 3 longest-substring-without-repeating-characters
Topic link: https: //leetcode.com/problems/longest-substring-without-repeating-characters/
Analysis of Algorithms:
int lengthOfLongestSubstring(string s) { if(s.length() == 0) return 0; int result = 0; vector<char> temp_chars; temp_chars.push_back(s[0]); for(int i = 1; i < (int)s.length(); i++){ vector<char>::iterator iter = find(temp_chars.begin(),temp_chars.end(),s[i]); if(iter != temp_chars.end()){ if(result < temp_chars.size()) result = temp_chars.size(); while(temp_chars[0] != s[i]){ temp_chars.erase(temp_chars.begin()); } temp_chars.erase(temp_chars.begin()); } temp_chars.push_back(s[i]); //showchars(temp_chars); } if(result < temp_chars.size()) result = temp_chars.size(); return result; }