Leetcode Note

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：

1. 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;
		}
	}

1. 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] == 2
   
   
   
   Runtime: 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:

1. (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$O(nlogn)$.
2. （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;
	}
};