第6章:LeetCode--数组(冒泡排序、快速排序)
时间:2019-08-29
本文章向大家介绍第6章:LeetCode--数组(冒泡排序、快速排序),主要包括第6章:LeetCode--数组(冒泡排序、快速排序)使用实例、应用技巧、基本知识点总结和需要注意事项,具有一定的参考价值,需要的朋友可以参考一下。
11. Container With Most Water
class Solution { public: int maxArea(vector<int>& height) { int maxarea = 0; int l = 0, r = height.size()-1; if(height.size()<2) return maxarea; while(l<r){ maxarea = max(maxarea, min(height[l],height[r])*(r-l)); if(height[l]<height[r]) l++; else r--; } return maxarea; } }; //Brute Force --Time Limit Exceed /*class Solution { public: int maxArea(vector<int>& height) { int maxarea = 0; if(height.size()<2) return maxarea; for(int i=0; i<height.size()-1; i++){ for(int j=i+1; j<height.size(); j++) maxarea = max(maxarea, min(height[i],height[j])*(j-i)); } return maxarea; } };*/
16. 3Sum Closest //最接近target的 三数和
class Solution { public: int threeSumClosest(vector<int>& nums, int target) { if(nums.size()<3) return 0; if(nums.size() == 3) return nums[0]+nums[1]+nums[2]; sort(nums.begin(), nums.end()); //int diff = abs(nums[0]+nums[1]+nums[2]-target); int diff = INT_MAX; //int sum = 0, front, end, ret=diff; int sum = 0, front, end, ret=0; for(int i=0; i<nums.size(); i++){ front = i+1; end = nums.size()-1; while(front<end){ sum = nums[i]+nums[front]+nums[end]; if(sum == target) return target; if(abs(sum-target)<diff){ diff = abs(sum-target); ret = sum; } (sum > target) ? end-- : front++; } } return ret; } };
49. Group Anagrams
Input: ["eat", "tea", "tan", "ate", "nat", "bat"], Output: [ ["ate","eat","tea"], ["nat","tan"], ["bat"] ]
class Solution { public: vector<vector<string>> groupAnagrams(vector<string>& strs) { unordered_map<string, vector<string>> mp; for (string s : strs) { //auto s: strs also ok string t = s; sort(t.begin(), t.end()); mp[t].push_back(s); } vector<vector<string>> anagrams; for (auto p : mp) { anagrams.push_back(p.second); } return anagrams; } };
819. Most Common Word
Input: paragraph = "Bob hit a ball, the hit BALL flew far after it was hit." banned = ["hit"] Output: "ball" class Solution { public: string mostCommonWord(string paragraph, vector<string>& banned) { // Map out each banned word // Go through each word, if it isn't in the banned list // put it in another map and keep track of how many times it appears // Keep two variables to maintain the mostCommonWord and how many times it appeared // This prevents us from looping thorugh the map at the end to find the most common word map<string, int> bannedWords; map<string, int> tracker; string mostCommonWord = ""; int mostCommonWordCount = 0; for(int x = 0; x < banned.size(); x++) { bannedWords[banned[x]] = 0; } for(int x = 0; x < paragraph.length(); x++) { string temp = ""; while(x < paragraph.length() && paragraph[x] != ' ') { if(!isalpha(paragraph[x])) { break; } temp += tolower(paragraph[x]); x++; } if(temp == " " || temp == "") continue; if(bannedWords.count(temp) != 0) continue; else { tracker[temp]++; if(mostCommonWordCount < tracker[temp]) { mostCommonWordCount = tracker[temp]; mostCommonWord = temp; } } } return mostCommonWord; } };
Question1???:
Input: inputStr = awaglknagawunagwkwagl num = 4 Output: {wagl, aglk, glkn, lkna, knag, gawu, awun, wuna, unag, nagw, agwk, kwag}
Explanation: Substrings in order are: wagl, aglk, glkn, lkna, knag, gawu, awun, wuna, unag, nagw, agwk, kwag, wagl "wagl" is repeated twice, but is included in the output once.
class Solution { public: vector<string> numKLenSubstrNoRepeats(string S, int K) { vector<string>ans; int cnt = 0, ,i=0, j; while(i < S.size() && j<k) { if(S[i] == S[j]){ j++; }else{ i = i-j+1; j = 0; } if (j == K){ ans.push_back(S.substr(i-K, K)); i = i-j+1; j = 0; } } return ans; } };
数组排序
https://www.cnblogs.com/taotingkai/p/6214367.html
/* 冒泡排序 */ /* 1. 从当前元素起,向后依次比较每一对相邻元素,若逆序则交换 */ (每次把最大值放后面) /* 2. 对所有元素均重复以上步骤,直至最后一个元素 */ void bubbleSort (int arr[], int len) { int temp; int i, j; for (i=0; i<len-1; i++) /* 外循环为排序趟数,len个数进行len-1趟 */ for (j=0; j<len-1-i; j++) { /* 内循环为每趟比较的次数,第i趟比较len-i次 */ if (arr[j] > arr[j+1]) { /* 相邻元素比较,若逆序则交换(升序为左大于右,降序反之) */ temp = arr[j]; arr[j] = arr[j+1]; arr[j+1] = temp; } } } /* 每次选出最小的 */ void bubbleSort (int arr[], int len) { int temp; int i, j; for (i=0; i<len-1; i++) /* 外循环为排序趟数,len个数进行len-1趟 */ for (j=i+1; j<len-1; j++) { /* 内循环为每趟比较的次数,第i趟比较len-i次 */ if (arr[i] > arr[j]) { /* 相邻元素比较,若逆序则交换(升序为左大于右,降序反之) */ temp = arr[i]; arr[i] = arr[j]; arr[j] = temp; } } }
/*快速排序*/
#include <iostream> using namespace std; void Qsort(int arr[], int low, int high){ if (high <= low) return; int i = low; int j = high + 1; int key = arr[low]; while (true) { /*从左向右找比key大的值*/ while (arr[++i] < key) { if (i == high){ break; } } /*从右向左找比key小的值*/ while (arr[--j] > key) { if (j == low){ break; } } if (i >= j) break; /*交换i,j对应的值*/ int temp = arr[i]; arr[i] = arr[j]; arr[j] = temp; } /*中枢值与j对应值交换*/ int temp = arr[low]; arr[low] = arr[j]; arr[j] = temp; Qsort(arr, low, j - 1); Qsort(arr, j + 1, high); } int main() { int a[] = {6, 2, 7, 3, 8, 9}; Qsort(a, 0, sizeof(a) / sizeof(a[0]) - 1);/*这里原文第三个参数要减1否则内存越界*/ for(int i = 0; i < sizeof(a) / sizeof(a[0]); i++) { cout << a[i] << ""; } return 0; }/*参考数据结构p274(清华大学出版社,严蔚敏)*/
求一个数组里面连续子集的和的最大值,比如[1,2,-4,7,8,-2,4]=>[7,8,-2,4]
思路:每次求出max(sub(i-1), arr[i]),看那个大取那个。
int arrMaxsub(int [] arr, int len){
int maxsum = arr[0]; for(int i=1; i<len; i++){ if(maxsum+arr[i] <arr[i]) maxsum = arr[i]; else maxsum = maxsum+arr[i]; }
return maxsum; }
原文地址:https://www.cnblogs.com/feliz/p/11145194.html
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