第1章:LeetCode--基础部分
2019-08-29 08:56:07来源:博客园 阅读 ()
第1章:LeetCode--基础部分
LeetCode刷题指导(不能只实现功能就行需要尽量写最优解):
不可能刷遍所有题,只能通过刷题来恢复写代码的功力,熟悉常用算法(暴力算法、冒泡排序/快速排序、strStr KMP算法、二叉树遍历DFS/BFS算法、二叉树前序遍历/中序遍历/后序遍历算法),以及一些常考题目(链表反转、快慢指针、链表插入删除)等。
可以先看TOP100里面的easy和medium的(本篇基础部分(1)),然后再按数组、链表、字符串类刷easy和medium的(参看后续篇章),大概刷完4,50道题后,要把题目归类总结,打印出来,经常看。
下面有一些刷题顺序的例子也可以参考:https://blog.csdn.net/love1055259415/article/details/80981337
#include <map> #include <iostream> #include <string> #include <stack> #include <queque> using namespace std; //1.two-sum //C int* twoSum(int* nums, int numSize, int target, int* returnSize){ for(int i=0; i<numSize; i++){ for(int j=i+1; j<numSize; j++){ if(target-numSize(i) == nums[j]){ *returnSize = 2; int* ret = (int*)malloc(2*sizeof(int)); ret[0] = i; ret[1] = j; return ret; } } } *returnSize = 0; return NULL; } //C++ vector<int> twoSum(vector<int>& nums, int target){ unordered_map<int, int> mymap; for(int i=0; i<nums.size(); i++){ if(mymap.find(target-nums[i]) != mymap.end()) return {mymap[target – nums[i]], i}; mymap[nums[i]] = i; } return {}; } // 2. Add Two Numbers /** * 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) { int sum=0, carry=0; ListNode *ret, *tmp, *p=l1, *q=l2; ret = tmp = new ListNode(0); while (carry || p || q){ //如果没有carry, [5] /[5] 时就会得[0,0] sum = carry; if(p) sum += p->val; if(q) sum += q->val; tmp->val = sum%10; carry = sum/10; p = (p && p->next)? p->next : NULL; //如果不判断[1,8] / [0] 会出错 q = (q && q->next)? q->next : NULL; if( !carry && !p && !q) break;//如果没有这个[7,0,8,0] tmp->next = new ListNode(0); tmp = tmp->next; } return ret; } }; //3. Longest Substring Without Repeating Characters class Solution { //abcdabcef public: int lengthOfLongestSubstring(string s) { vector<int> dict(256, -1); int maxLen = 0, start = -1; for (int i = 0; i < s.length(); i++) { if (dict[s[i]] > start) //重复出现了字符,更改start位置 start = dict[s[i]]; dict[s[i]] = i; //更新字符对应的下标 maxLen = max(maxLen, i - start ); //返回最大值i-start } return maxLen; } }; //4.Median of two sorted Arrays double findMedianSortedArrays(vector<int>& num1, vector<int>& num2){ vector<int> nums(nums1); nums.insert(nums.end(), num2.begin(), num2.end()); sort(nums.begin(), nums.end()); int size = nums.size(); if(size%2 == 0) return (nums[size/2-1]+nums[size/2])/2.0; //(nums[size/2-1]+nums(size/2))/2. will not get xx.5 else return nums[size/2]; } //A[i] B[j], i+j=(n+m+1)/2 Dont use STL, find the (m+n)/2 data: class Solution { public: double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) { int m = nums1.size(); int n = nums2.size(); int p1=0, p2=0; int count=0, loop=(m+n)/2+1; int value=0, pre=0; while(true){ value = 0; if(p1<m && p2<n){ if(nums1[p1]<nums2[p2]) value = nums1[p1++]; else value = nums2[p2++]; }else if(p1>=m && p2<n){ value = nums2[p2++]; }else if(p1<m && p2>=n){ value = nums1[p1++]; }else break; count++; if(count == loop){ if((m+n)%2) return value; else return (pre+value)/2.0; } pre = value; } return 0; } }; // 5. Longest Palindromic Substring “abbab” ->abba string longestPalindrome(string s){ int len = s.size(); for(int sublen=len; sublen>0; sublen--){ //可能的最大串长度 int i=0, j=0; //开始判断是否有这么长的palindromic串 while(i+sublen<=len){ j = 0; int loop = sublen/2; while(j<loop){ if(s[i+j] != s[i+sublen-1-j]) //sublen长的串从最两端向里判断 break; j++; } if(j == loop) return s.substr(i,sublen); i+=1; //没找到,前进一个字符 } } return s; } // 6. ZigZag Conversion字母变成|/|/|/这种排列 class Solution { public: string convert(string s, int numRows) { vector< vector<char> > str(numRows); //must give the lines, or Line 933: Char 34: runtime error: reference binding to null pointer if(s.size()<2) return s; if(numRows<2) return s; int strLen = s.size(), curr=0; string rets=""; while(curr<strLen){ for(int i=0; i<numRows; i++){ //”|“/ fill the colume if(curr<strLen) str[i].push_back(s[curr++]); else break; } for(int i=numRows-2; i>0; i--){ // |”/” fill the other lines if(curr<strLen) str[i].push_back(s[curr++]); else break; } } int i=0; while(i<numRows){ for(int j=0; j<str[i].size(); j++) rets += str[i][j]; i++; } return rets; } }; // 7. Reverse Integer 123-321 -123 -321 120 12 int reverse(int x){ Int ret=0, div=0; while(x){ div = x%10; x = x/10; ret = ret*10 + div; } return ret; } // 8. String to Integer (atoi) class Solution { public: int myAtoi(string str) { if(str.empty()) return 0; long retValue = 0, j=0, minusFlag=1, index=0, terminal=0; while(str[j]){ switch(str[j]){ case ' ': if(terminal) return retValue; break; case '+': if(terminal) return retValue; if(++index > 1) return 0; terminal =1; minusFlag=1; break; case '-': if(terminal) return retValue; if(++index > 1) return 0; terminal =1; minusFlag=-1; break; case '0': case '1': case '2': case '3': case '4': case '5': case '6': case '7': case '8': case '9': terminal =1; retValue = retValue*10 + minusFlag*((str[j]-'0')); if(retValue<INT_MIN) return INT_MIN; if(retValue>INT_MAX) return INT_MAX; break; default: return retValue; } j++; } return retValue; } }; // 9. Palindrome Number 回文 int revert(int x){ Int ret = 0; while(x){ If(abs(ret)>INT_MAX/10) return 0; ret = ret*10+x%10; x=x/10; } return ret; } bool isPalindrome(int x){ if(x<0) return false; int reverse = 0; reverse = revert(x); if(reverse == x) return true; else return false; } // 13. Roman to Integer Symbol Value I 1 V 5 X 10 L 50 C 100 D 500 M 1000 • I can be placed before V (5) and X (10) to make 4 and 9. • X can be placed before L (50) and C (100) to make 40 and 90. • C can be placed before D (500) and M (1000) to make 400 and 900. class Solution { public: int romanToInt(string s) { int ret=0; char *p =&s[0]; while(*p) { switch (*p){ case 'I': if(*(p+1) == 'V') { ret += 4; *p++; } else if(*(p+1) == 'X') { ret += 9; *p++; } else ret += 1; break; case 'V': ret += 5; break; case 'X': if(*(p+1) == 'L') { ret += 40; *p++; } else if(*(p+1) == 'C') { ret += 90; *p++; } else ret += 10; break; case 'L': ret += 50; break; case 'C': if(*(p+1) == 'D') { ret += 400; *p++; } else if(*(p+1) == 'M') { ret += 900; *p++; } else ret += 100; break; case 'D': ret += 500; break; case 'M': ret += 1000; break; default: break; } p=p+1; } return ret; } }; // 14. Longest Common Prefix string longestCommonPrefix(vector<string>& strs){ if(strs.empty())return “”; string rets = “”; for(int i=0; i<strs[0].size();i++){ for(int j=0; j<strs.size();j++){ if(strs[0][i] != strs[j][i]) return rets; } rets += strs[0][i]; } return rets; } ///////////////////////// /*if(strs.empty()) return ""; for(int idx = 0; strs[0][idx] != '\0'; ++idx) { for(auto& str : strs ) if(str[idx] != strs[0][idx]) return strs[0].substr(0,idx); } return strs[0];*/ ///////////////////////// //15. 3Sum vector<vector<int>> threeSum(vector<int>& nums) { vector<vector<int>> ret; std::sort(nums.begin(), nums.end()); // Sort the nums int size = nums.size(), a, b, c, front, end; for(int i = 0; i<size-2; i++){ a = nums[i]; front = i+1; end = size-1; while(front < end){ if(nums[front]+nums[end]+a < 0) front++; else if(nums[front]+nums[end]+a > 0) end--; else{ vector<int> elem({a, nums[front], nums[end]}); ret.push_back(elem); while(front<end && nums[front]==elem[1]) front++;//remove the duplicated nums. while(front<end && nums[end]==elem[2]) end--;//remove the duplicated nums. } } // Processing duplicates of Numbers while (i + 1 < nums.size() && nums[i + 1] == nums[i]) i++; } return ret; } // 20. Valid Parentheses '(', ')', '{', '}', '[' and ']' , determine if the input string is valid. bool isValid(string s) { vector<char> OpStr; int len = s.length(); char ee; for(int i=0; i<len; i++){ switch (s[i]){ case '(': case '[': case '{': OpStr.push_back(s[i]); break; case ')': if(OpStr.empty()) return false; ee = OpStr.back(); if(ee == '(') OpStr.pop_back(); else return false; break; case ']': if(OpStr.empty()) return false; ee = *(OpStr.end()-1); if(ee == '[') OpStr.pop_back(); else return false; break; case '}': if(OpStr.empty()) return false; ee = *(OpStr.rbegin()); if(ee == '{') OpStr.pop_back(); else return false; break; default: return false; } } if(OpStr.empty()) return true; else return false; } // 21. Merge Two Sorted Lists ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) { if(!l1) return l2; if(!l2) return l1; ListNode *head, *retList; if(l1->val <l2->val){ retList = head = l1; l1 = l1->next; } else{ retList = head = l2; l2 = l2->next; } while(l1 && l2){ if(l1->val <l2->val){ head->next = l1; l1 = l1->next; head = head->next; head->next = NULL; } else{ head->next = l2; l2 = l2->next; head = head->next; head->next = NULL; } } if(!l1) head->next = l2; if(!l2) head->next = l1; return retList; } // 26. Remove Duplicates from Sorted Array [1,1,2], int removeDuplicates(vector<int>& nums) { if(nums.empty()) return 0; int i=0, j=1, size = nums.size(); for( ;j<size; j++){ if(nums[i] != nums[j]) nums[++i] = nums[j]; } return i+1; } int removeDuplicates(vector<int>& nums) { if(nums.empty()) return 0; vector<int>::iterator it; for(it=nums.begin(); it<nums.end()-1; it++){ if(*it == *(it+1)){ nums.erase(it--); } } return nums.size(); } // 27. Remove Element [0,1,2,2,3,0,4,2], val = 2, return length = 5: 0, 1, 3, 0, and 4. int removeElement(vector<int>& nums, int val) { if(nums.empty()) return 0; int i=0, j=0; for(int j=0; j<nums.size(); j++){ if(nums[j] != val) nums[i++] = nums[j]; } return i; } int removeElement(vector<int>& nums, int val) { if(nums.empty()) return 0; vector<int>::iterator it; for(it=nums.begin(); it<nums.end(); it++){ if(*it == val) nums.erase(it--); } return nums.size(); } // 28. Implement strStr() 参考KMP算法 // 35. Search Insert Position Input: [1,3,5,6], 5 Output: 2 int searchInsert(vector<int>& nums, int target) { int size = nums.size(), i=0; for(; i<size; i++){ if(nums[i] <target) continue; if(nums[i] >= target) return i; } return i; } // 38. Count and Say 1. 1 2. 11 3. 21 4. 1211 5. 111221 string countAndSay(int n) { string res, tmp; if (n == 1) return "1"; while (n>0){ int count = 1; res = countAndSay(--n); tmp = ""; for (int i = 0; i<res.size(); i++){ if (res[i] == res[i + 1]) count++; else{ tmp += to_string(count) + res[i]; count = 1; } } return tmp; } return tmp; } string countAndSay(int n) { if(n == 1)return "1"; string ret="", s1 = "1", currS= s1; int count = 1; for(int i=2; i<=n; i++){ ret = ""; for(int j=0; j<currS.size(); j++){ if(currS[j] == currS[j+1]) count++; else{ ret += to_string(count) + currS[j]; count = 1; } } count = 1; currS = ret; } return ret; } // 56. Merge Intervals Input: [[1,3],[2,6],[8,10],[15,18]] Output: [[1,6],[8,10],[15,18]] vector<vector<int>> merge(vector<vector<int>>& intervals){ int len = intervals.size(); if(len<2) return intervals; //len=0,1 return sort(intervals.begin(), intervals.end()); int j = 1; vector<vector<int>> ret; ret.push_back(intervals[0]); for(;j<len;j++){ if(ret.back()[1]<intervals[j][0]) ret.push_back(intervals[j]); else if(ret.back()[1]<intervals[j][1]) ret.back()[1]=intervals[j][1]; } return ret; } // 57. Insert Interval Input: intervals = [[1,3],[6,9]], newInterval = [2,5] Output: [[1,5],[6,9]] vector<vector<int>> insert(vector<vector<int>>& intervals, vector<int>& newInterval) { vector<vector<int>> ret, merged(intervals); if(!newInterval.empty()) merged.push_back(newInterval); sort(merged.begin(), merged.end()); int len = merged.size(); if(len<=1) return merged; ret.push_back(merged[0]); for(int j=1; j<len; j++){ if(ret.back()[1]<merged[j][0]) ret.push_back(merged[j]); else ret.back()[1] = max(ret.back()[1],merged[j][1]); } return ret; }
原文链接:https://www.cnblogs.com/feliz/p/11424903.html
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