算法刷题笔记

Leetcode-11. Container With Most Water

Method: （对撞指针）每次保留两指针中最大的那个即可求得最大的面积Runtime: 16 ms, faster than 95.65% of C++ online submissions for Container With Most Water.Memory Usage: 9.9 MB, less than 43.30% of C++ online submissions for Container With Most Water.

class Solution { public: int maxArea(vector<int>& height) { int i = 0; int r = height.size()-1; long long max = -1; while (i<r) { long long area = (r - i)*min(height[r], height[i]); if (area > max) max = area; if (height[r] > height[i]) i++; else r--; } return max; } };

Leetcode-26 Remove Duplicates from Sorted Array

Method: temp 作为一个标志表示遍历过的数，用temp来比较是否重复。Runtime: 16 ms, faster than 99.08% of C++ online submissions for Remove Duplicates from Sorted Array.Memory Usage: 9.8 MB, less than 97.50% of C++ online submissions for Remove Duplicates from Sorted Array.

int removeDuplicates(vector<int>& nums) { int length = 0; int temp = 99999; for (int i = 0;i < nums.size();i++) { if (nums[i] != temp) { nums[length++] = nums[i]; temp = nums[i]; } } return length; }

Leetcode-27 Remove Element

Method: k为非val值的个数, 查询到一个重复值$‘i’$则继续往下遍历，若非val值则进行赋值。Runtime: 0 ms, faster than 100.00% of C++ online submissions for Remove Element.Memory Usage: 8.5 MB, less than 97.06% of C++ online submissions for Remove Element.

int removeElement(vector<int>& nums, int val) { int k = 0; int m = 0; for (int i = 0;i < nums.size();i++) { if (nums[i] == val) { m++; } else { if (k != m) { nums[k] = nums[m]; } k++;m++; } } return k; }

Leetcode-75 Sort Color

Method：

可以采用计数排序，统计0，1，2个数然后按数量插入就好。Runtime: 4 ms, faster than 68.84% of C++ online submissions for Sort Colors.Memory Usage: 8.5 MB, less than 98.25% of C++ online submissions for Sort Colors. void sortColors(vector<int>& nums) { const int n = 3; int count[n] = { 0 }; for (int i = 0;i < nums.size();i++) { count[nums[i]]++; } int index = 0; for (int i = 0;i < n;i++) { for (int j = 0;j < count[i];j++) nums[index++] = i; } } 三路快排方法实现，只需要执行一次三路快排,遍历一遍把输入序列分成以下三块：arr[0 ... zero] == 0 / arr[zero+1 ... i-1] == 1 / arr[two ... n-1] == 2Runtime: 0 ms, faster than 100.00% of C++ online submissions for Sort Colors.Memory Usage: 8.5 MB, less than 96.49% of C++ online submissions for Sort Colors void sortColors(vector<int>& nums) { int zero = -1; //nums[0 .. zero] == 0 int two = nums.size(); // nums[two..-n-1] == 2 for (int i = 0;i < two;) { if (nums[i] == 1) i++; else if (nums[i] == 2) { two--; swap(nums[i], nums[two]); } else {// nums[i] == 0 zero++; swap(nums[zero], nums[i]); i++; } } }

Leetcode-80 Remove Duplicates from Sorted Array II

Method: @k 为合法数字的数量，@t 判断一类数字是否<2， @temp 用于比较的数字的拷贝， 迭代位置与合法数字的位置做交换，一次遍历即可完成。 Runtime: 8 ms, faster than 99.12% of C++ online submissions for Remove Duplicates from Sorted Array II. Memory Usage: 8.8 MB, less than 89.47% of C++ online submissions for Remove Duplicates from Sorted Array II.

int removeDuplicates(vector<int>& nums) { int k = 0; bool t = true; int temp = 99999; for (int i = 0;i < nums.size();i++) { if (nums[i] == temp) { if (t == true) continue; else { t = true; if (k != i) { nums[k] = nums[i]; } k++; } } else { t = false; temp = nums[i]; if (k != i) { nums[k] = nums[i]; } k++; } } return k; }

Leetcode-88 Merge Sorted Array

Method: 归并排序，设置一个外置数组用以空间换时间，时间为遍历一遍+拷贝长数组一遍的时间。 Runtime: 0 ms, faster than 100.00% of C++ online submissions for Merge Sorted Array. Memory Usage: 8.7 MB, less than 80.43% of C++ online submissions for Merge Sorted Array.、

void merge(vector<int>& nums1, int m, vector<int>& nums2, int n) { int p1 = 0, p2 = 0; vector<int> temp; for (int i = 0;i < m;i++) { temp.push_back(nums1[i]); } int index = 0; while (p1 < m && p2 < n) { if (temp[p1] <= nums2[p2]) { if (index != p1) nums1[index] = temp[p1]; index++; p1++; } else { nums1[index++] = nums2[p2++]; } } if (p1 != m) for (;p1 < m;p1++) nums1[index++] = temp[p1]; else for (;p2 < n;p2++) nums1[index++] = nums2[p2]; }

Leetcode-125. Valid Palindrome

Method: （对撞指针）先判断字符的有效性，再判断是否要转换大小写，最后在相同类型的文字格式下进行字符的比较。 Runtime: 12 ms, faster than 39.15% of C++ online submissions for Valid Palindrome. Memory Usage: 9.3 MB, less than 81.63% of C++ online submissions for Valid Palindrome.

bool isPalindrome(string s) { int p = 0; int r = s.length() - 1; while (p<r) { if ((s[p] >= 'A'&& s[p] <= 'Z') || (s[p] >= 'a'&&s[p] <= 'z') || (s[p] >= '0'&&s[p] <= '9')) { if ((s[r] >= 'A'&& s[r] <= 'Z') || (s[r] >= 'a'&&s[r] <= 'z') || (s[r] >= '0'&&s[r] <= '9')) { char c = s[p]<'A'?s[p]:(s[p] >= 'a' ? s[p] : s[p] - 'A' + 'a'); char k = s[r]<'A' ? s[r] : (s[r] >= 'a' ? s[r] : s[r] - 'A' + 'a'); if (c != k) { return false; } else { p++; r--; } } else { r--; continue; } } else { p++; } } return true; }

Leetcode-167 Two Sum II - Input array is sorted(*)Method:

(二分查找)从起始点开始遍历，在除了遍历元素$num[i]$外的有序数组结构中采用二分搜索的方法查找$target-nums[i]$可以使时间复杂度控制在$O(nlogn)$.（对撞指针）前后各一个位置的指针，计算 $ nums[i] + nums[j] < target ——> i++ $ $ nums[i] + nums[j] > target ——> j++ $ 只需要一次遍历即可找到答案 Runtime: 4 ms, faster than 96.98% of C++ online submissions for Two Sum II - Input array is sorted. Memory Usage: 9.5 MB, less than 82.35% of C++ online submissions for Two Sum II - Input array is sorted. vector<int> twoSum(vector<int>& numbers, int target) { assert(numbers.size() >= 2); int l = 0, r = numbers.size() - 1; while (l < r) { if (numbers[l] + numbers[r] == target) { int res[2] = { l + 1, r + 1 }; return vector<int>(res, res + 2); } else if (numbers[l] + numbers[r] < target) l++; else r--; //numbers[l] + numbers[r] > target } throw invalid_argument("The input has no solution."); }

Leetcode-215. Kth Largest Element in an Array()Method: （快速排序）运用快排partition操作 返回标定点位置，我们能够知道，标定点的位置是每一轮快排过后确定的最终位置因此只要 标定点位置p与想要查找的位置k相重则说明标定点已经排序完成，只需要返回标定点在数组中的值就好了。

Runtime: 28 ms, faster than 29.31% of C++ online submissions for Kth Largest Element in an Array.Memory Usage: 11 MB, less than 6.06% of C++ online submissions for Kth Largest Element in an Array.

int _partition(vector<int>& arr, int l, int r) { int temp = arr[l]; int i = l + 1, j = r; while (true) { while (i <= r && arr[i] > temp)i++; while (j >= l+1 && arr[j] < temp)j--; if (i > j) break; swap(arr[i], arr[j]); i++; j--; } swap(arr[l], arr[j]); return j; } int __quickSort(vector<int>& arr, int l, int r, int k) { int p = _partition(arr, l, r); if (p != k-1) { if(p > k-1) return __quickSort(arr, l, p - 1, k); else return __quickSort(arr, p + 1, r, k); } else return arr[p]; } int findKthLargest(vector<int>& nums, int k) { return __quickSort(nums, 0, nums.size()-1, k); }

Leetcode-283 Move Zeroes

Method:主要用了一个标定点** $i$ 来标识一轮遍历过程中的非零数的个数，然后将每个非零数按$i$的位置存放。时间复杂度$O(n)$；空间复杂度$O(1)$。Others Method：** 使用交换思想，只需要一轮遍历.Runtime: 12 ms, faster than 96.77% of C++ online submissions for Move Zeroes.Memory Usage: 9.3 MB, less than 100.00% of C++ online submissions for Move Zeroes.

void moveZeroes(vector<int>& nums) { int i = 0; for (int j = 0;j < nums.size();j++) { if (nums[j] != 0) { nums[i++] = nums[j]; } } for (;i < nums.size();i++) { nums[i] = 0; } } //交换方法 void moveZeroes(vector<int>& nums) { int i = 0; for (int j = 0;j < nums.size();j++) { if (nums[j] != 0) { if (i != j) { int temp = nums[i]; nums[i++] = nums[j]; nums[j] = temp; } else //i==k i++; } } }

Leetcode 344. Reverse String

Method:(对撞指针)简单用法。 Runtime: 44 ms, faster than 92.20% of C++ online submissions for Reverse String. Memory Usage: 15.2 MB, less than 86.59% of C++ online submissions for Reverse String.

void reverseString(vector<char>& s) { int p = 0; int r = s.size()-1; while (p < r) { swap(s[p++], s[r--]); } }

Leetcode 345. Reverse Vowels of a String

Method:(对撞指针)简单用法。 Runtime: 4 ms, faster than 99.53% of C++ online submissions for Reverse Vowels of a String. Memory Usage: 10 MB, less than 87.88% of C++ online submissions for Reverse Vowels of a String

class Solution { public: string reverseVowels(string s) { int p = 0; int r = s.length()-1; while (p < r) { if (s[p] == 'a' || s[p] == 'A' || s[p] == 'e' || s[p] == 'E' || s[p] == 'i' || s[p] == 'I' || s[p] == 'o' || s[p] == 'O' || s[p] == 'u' || s[p] == 'U') if (s[r] == 'a' || s[r] == 'A' || s[r] == 'e' || s[r] == 'E' || s[r] == 'i' || s[r] == 'I' || s[r] == 'o' || s[r] == 'O' || s[r] == 'u' || s[r] == 'U') swap(s[p++], s[r--]); else r--; else p++; } return s; } };

转载于:https://www.cnblogs.com/simon-slrn/p/11386553.html

相关资源：Leetcode_Cpp：leetcode_note-源码