Leetcode Problem 汇总二

最新推荐文章于 2024-10-08 16:45:42 发布

原创最新推荐文章于 2024-10-08 16:45:42 发布 · 1.2k 阅读

1 ·

CC 4.0 BY-SA版权

本文精选了LeetCode上的二十道经典算法题，包括跳格子、组合问题、链表操作、路径求和等，提供了详细的解题思路及C++、Python代码实现，帮助读者深入理解算法原理。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

第一题：
题目：

Jump Game II

地址：http://oj.leetcode.com/problems/jump-game-ii/

分析：O(n)就可以。DP问题。

代码：

(Python版)

class Solution:
    # @param A, a list of integers
    # @return an integer
    def jump(self, A):
        l1 = len(A)
        right = A[0]
        if l1 == 1:return 0
        if right >= l1-1: return 1
        if right == 0: ans = 0
        else: ans = 1
        maxnum = 0
        i = 1
        while i < l1:
            if i <= right:
                if i+A[i] > maxnum:
                    maxnum = i+A[i]
                i += 1
            else:
                if maxnum == right: right += 1
                else: right = maxnum
                maxnum = 0
                ans += 1
                if right >= l1-1:
                    return ans

第二题：

Combinations

地址：http://oj.leetcode.com/problems/combinations/

分析：排列问题，DFS

代码：

(C++版)

class Solution 
{
public:
	int arr[1000];
	int visit[1000];
	int ar[1000];
	int K;
	int N;
	vector<vector<int>> ans;
	void dfs(int count, int pos)
	{
		int i;
		if(count > K)
		{
			vector<int> add;
			for(i = 1; i < count; i++)
				add.push_back(ar[i]);
			ans.push_back(add);
			return ;
		}
		for(i = pos; i <= N; i++)
		{
			if(visit[i] == 0)
			{
				visit[i] = 1;
				ar[count] = i;
				dfs(count+1, i+1);
				visit[i] = 0;
			}
		}
	}

    vector<vector<int> > combine(int n, int k) 
	{
		int i;
		for(i = 0; i < 1000; i++) arr[i] = i;
		for(i = 0; i < 1000; i++) visit[i] = 0;
		K = k;
		N = n;
		dfs(1,1);
		return ans;
    }
};

第三题：

Swap Nodes in Pairs

地址：http://oj.leetcode.com/problems/swap-nodes-in-pairs/

分析：指针题，指针用的不熟练，有点DT。。。

代码：

(C++版)

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode *swapPairs(ListNode *head) 
	{
		int i;
		int flag = 0;
		if(head == NULL) return head;
		ListNode *go, *ans, *pre;
		ans = (ListNode *)malloc(sizeof(ListNode));
		ans->next = head;
		pre = ans;
		while(head != NULL)
		{
		    if(head->next != NULL)
		    {
			    go = head->next->next;
			    head->next->next = head;
				pre->next = head->next;
				head->next = go;
				pre = head;
			    head = go;
		    }
		    else break;
		}
		return ans->next;
    }
};

第四题：

Remove Duplicates from Sorted List

地址：http://oj.leetcode.com/problems/remove-duplicates-from-sorted-list/

分析：指针链表的应用，基础题。注意最后head->next = NULL;

代码：

(C++版)

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution 
{
public:
    ListNode *deleteDuplicates(ListNode *head) 
	{
	    if(head == NULL) return NULL;
        ListNode *ans, *go;
		ans = (ListNode *)malloc(sizeof(ListNode));
		ans->next = head;
		go = head->next;
		if(go == NULL) return ans->next;
		while(go != NULL)
		{
			if(head->val == go->val)
				go = go->next;
			else
			{
				head->next = go;
				head = head->next;
				go = go->next;
			}
		}
		head->next = NULL;

		return ans->next;
    }
};

第五题：

Two Sum

地址：http://oj.leetcode.com/problems/two-sum/

分析：记录下index，把value排序，然后利用two pointers就OK了。

代码：

(C++版)

typedef struct Node
{
	int value, index;
}Node;

class Solution 
{
public:
	static bool cmp(const Node &a, const Node &b)
	{
		return a.value < b.value;
	}
    vector<int> twoSum(vector<int> &numbers, int target) 
	{
		vector<int> ans;
		vector<Node> nn;
		int size = numbers.size();
		int i, j;
		for(i = 0; i < size; i++)
		{
			Node add;
			add.value = numbers[i];
			add.index = i;
			nn.push_back(add);
		}
		sort(nn.begin(), nn.end(), cmp);
		i = 0, j = size-1;
		while(i < j)
		{
			if(nn[i].value+nn[j].value == target)
			{
				ans.push_back(nn[i].index+1);
				ans.push_back(nn[j].index+1);
				break;
			}
			else if(nn[i].value+nn[j].value > target)
				j--;
			else i++;
		}
		sort(ans.begin(), ans.end());
		//cout << ans[0] << " " << ans[1] << endl;
		return ans;
    }
};

第六题：

Path Sum

地址：http://oj.leetcode.com/problems/path-sum/

分析：递归。。。

代码：

(C++版)

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
int target;
class Solution 
{
public:
	static bool dfs(TreeNode *root, int count)
	{
		count += root->val;
		if(count == target && root->left == NULL && root->right == NULL) return true;
		if(root->left != NULL)
		{
			if(dfs(root->left, count) == true) return true;
		}
		if(root->right != NULL)
		{
			if(dfs(root->right, count) == true) return true;
		}
		return false;
	}
    bool hasPathSum(TreeNode *root, int sum)
	{
        bool flag = false;
		target = sum;
		if(root == NULL) return flag;
		flag = dfs(root, 0);
		return flag;
    }
};

第七题：

Path Sum II

地址：http://oj.leetcode.com/problems/path-sum-ii/

分析：递归。

代码：

(C++版)

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
int target;
vector<vector<int>> ans;
vector<int> add;
class Solution 
{
public:
	static void dfs(TreeNode *root, int count)
	{
		count += root->val;
		add.push_back(root->val);
		if(root->left == NULL && root->right == NULL)
		{
			if(count == target)
				ans.push_back(add);
			add.pop_back();
			return ;
		}
		if(root->left != NULL)
		{
			dfs(root->left, count);
		}
		if(root->right != NULL)
			dfs(root->right, count);
		add.pop_back();
		return ;
	}
    vector<vector<int> > pathSum(TreeNode *root, int sum) 
	{
		target = sum;
		if(root == NULL) return ans;
		ans.clear();
		add.clear();
		dfs(root, 0);
		
		return ans;
    }
};

第八题：

Single Number II

地址：http://oj.leetcode.com/problems/single-number-ii/

分析：类似于前面那个single number，转化为3进制。注意正数与负数。

代码：

(C++版)

int arr[1000];
int ar[1000];
class Solution 
{
public:
    int singleNumber(int A[], int n) 
	{
		int i, pos;
		if(n == 1) return A[0];
		for(i = 0;i < 1000; i++) arr[i] = ar[i] = 0;
        for(i = 0; i < n; i++)
		{
			if(A[i] >= 0)
			{
				pos = 0;
				long long int num = A[i];
				while(num > 0)
				{
					arr[pos] += num%3;
					arr[pos] %= 3;
					num /= 3;
					pos += 1;
				}
			}
			else
			{
				pos = 0;
				long long int num = (long long int)0-A[i];
				while(num > 0)
				{
					ar[pos] += num%3;
					ar[pos] %= 3;
					num /= 3;
					pos += 1;
				}
			}
		}

		for(i = 100;i >= 0; i--)
			if(arr[i] != 0) break;
		long long ans = 0;
		for( ; i >= 0; i--)
		{
			ans *= 3;
			ans += arr[i];
		}
		if(ans == 0)
		{
			for(i = 100; i >= 0; i--)
				if(ar[i] != 0) break;
			for(; i >= 0; i--)
			{
				ans *= 3;
				ans += ar[i];
			}
			ans = 0 - ans;
		}
		return ans;
    }
};

第九题：

Jump Game

地址：http://oj.leetcode.com/problems/jump-game/

分析：如上Jump Game II

代码：

(Python版)

class Solution:  
    # @param A, a list of integers  
    # @return an integer  
    def jump(self, A):  
        l1 = len(A)  
        right = A[0]  
        if l1 == 1:return 0  
        if right >= l1-1: return 1  
        if right == 0: ans = 0  
        else: ans = 1  
        maxnum = 0  
        i = 1  
        while i < l1:  
            if i <= right:  
                if i+A[i] > maxnum:  
                    maxnum = i+A[i]  
                i += 1  
            else:  
                if maxnum == right:
                    if right >= l1-1:
                        return True
                    else:
                        return False
                else: right = maxnum  
        return True

第十题：

Combination Sum II

地址：http://oj.leetcode.com/problems/combination-sum-ii/

分析：DFS

代码：

(C++版)

vector<vector<int>> ans;
vector<int> add;
int visit[1000];
class Solution 
{
public:
	vector<int> number;
	int len;
	static bool cmp(const vector<int> &a, const vector<int> &b)
	{
		int i;
		if(a.size() != b.size()) return a.size() < b.size();
		for(i = 0; i < a.size(); i++)
			if(a[i] != b[i]) return a[i] < b[i];
		return false;
	}
	void dfs(int sum, int pos, int target)
	{
		int i;
		if(sum == target)
		{
			ans.push_back(add);
			return;
		}
		for(i = pos; i < len; i++)
		{
			if(visit[i] == 0)
			{
				if(sum+number[i] <= target)
				{
					visit[i] = 1;
					add.push_back(number[i]);
					dfs(sum+number[i], i+1, target);
					add.pop_back();
					visit[i] = 0;
				}
			}
		}
	}

    vector<vector<int> > combinationSum2(vector<int> &num, int target) 
	{
		len = num.size();
		sort(num.begin(), num.end());
		int i, j;
		number = num;
		ans.clear();
		add.clear();
		for(i = 0; i < len; i++) visit[i] = 0;
		dfs(0, 0, target);
		vector<vector<int>> aaa;
		sort(ans.begin(), ans.end(), cmp);
		if(ans.size() == 0) return ans;
		int k = 0;
		aaa.push_back(ans[0]);
		for(i = 1; i < ans.size(); i++)
		{
			if(ans[i].size() == aaa[k].size())
			{
				int idx;
				for(idx = 0; idx < ans[i].size(); idx++)
				{
					if(ans[i][idx] != aaa[k][idx])break;
				}
				if(idx != ans[i].size())
				{
					aaa.push_back(ans[i]);
					k++;
				}
			}
			else
			{
				aaa.push_back(ans[i]);
				k++;
			}
		}

		return aaa;
    }
};

第十一题：

Combination Sum

地址：oj.leetcode.com/problems/combination-sum/

分析：DFS，比Sum II 简单不少。

代码：

(C++版)

vector<vector<int>> ans;
vector<int> add;
int size;
vector<int> number;
class Solution 
{
public:
	static void dfs(int pos, int sum, int target)
	{
		int i;
		if(sum == target)
		{
			ans.push_back(add);
			return;
		}
		for(i = pos; i < size; i++)
		{
			if(sum+number[i] <= target)
			{
				add.push_back(number[i]);
				dfs(i, sum+number[i], target);
				add.pop_back();
			}
		}
	}
    vector<vector<int> > combinationSum(vector<int> &candidates, int target) 
    {
		size = candidates.size();
		sort(candidates.begin(), candidates.end());
		number = candidates;
		add.clear();
		ans.clear();
		dfs(0, 0, target);
		
		return ans;
    }
};

第十二题：

Remove Duplicates from Sorted Array

地址：http://oj.leetcode.com/problems/remove-duplicates-from-sorted-array/

分析：水题。注意A为空时的处理。

代码：

(C++版)

class Solution 
{
public:
    int removeDuplicates(int A[], int n) 
	{
		int i, j = 1;
		int len = 1;
		if(n == 0) return 0;
		for(i = 1; i < n; i++)
			if(A[i] != A[i-1])
				A[j] = A[i], j++, len++;
		
		return len;
    }
};

第十三题

Remove Element

地址：http://oj.leetcode.com/problems/remove-element/

分析：水题

代码：

(C++版)

class Solution 
{
public:
    int removeElement(int A[], int n, int elem) 
	{
        int i, j;
		i = j = 0;
		if(n == 0) return 0;
		for(i = 0; i < n; i++)
		{
			if(A[i] == elem)
				j++;
			else
				A[i-j] = A[i];
		}
		return n - j;
    }
};

第十四题

Remove Duplicates from Sorted Array II

地址：http://oj.leetcode.com/problems/remove-duplicates-from-sorted-array-ii/

分析：哎。。。

代码：
(C++版)

class Solution
{
public:
    int removeDuplicates(int A[], int n)
    {
        int i, j = 1, k = 0;
        for(i = 1; i < n; i++)
        {
            if(A[i] == A[i-1])
            {
                j++;
                if (j > 2) k++;
            }
            else
                j = 1;
            A[i-k] = A[i];
        }
        return n-k;
    }
};

第十五题：

Unique Paths

地址：http://oj.leetcode.com/problems/unique-paths/

分析：机器人只能往右走或者往下走。简单DP。公式 dp[i][j] = dp[i][j-1] + dp[i-1][j];

代码：

(C++版)

int dp[110][110];
class Solution 
{
public:
    int uniquePaths(int m, int n) 
	{
		int i, j;
		for (i = 0; i < 110; i++) for (j = 0; j < 110; j++) dp[i][j] = 0;
		dp[0][1] = 1;
		for (i = 1; i <= m; i++)
			for (j = 1; j <= n; j++)
				dp[i][j] = dp[i-1][j] + dp[i][j-1];
		return dp[m][n];
    }
};

第十六题：

Unique Paths II

地址：http://oj.leetcode.com/problems/unique-paths-ii/

分析：如上题。只不过在grid[i][j] = 1时，dp[i+1][j+1] = 0;

代码：

(C++版)

int dp[110][110];
class Solution 
{
public:
	int uniquePathsWithObstacles(vector<vector<int> > &obstacleGrid) 
	{
		int m, n;
		int i, j;
		m = obstacleGrid.size();
		if (m == 0) return 0;
		n = obstacleGrid[0].size();
		memset(dp, 0, sizeof(dp));
		dp[0][1] = 1;
		for (i = 1; i <= m; i++)
		{
			for(j = 1; j <= n; j++)
			{
				if (obstacleGrid[i-1][j-1] == 1) dp[i][j] = 0;
				else dp[i][j] = dp[i-1][j] + dp[i][j-1];
			}
		}

		return dp[m][n];
	}
};

第十七题：

Binary Tree Preorder Traversal

地址：http://oj.leetcode.com/problems/binary-tree-preorder-traversal/

分析：PreOrder Travel

代码：

(C++版)

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
vector<int> ans;
class Solution 
{
public:
	static void dfs(TreeNode *root)
	{
		if (root == NULL) return ;
		ans.push_back(root->val);
		dfs(root->left);
		dfs(root->right);
	}
    vector<int> preorderTraversal(TreeNode *root) 
	{
		ans.clear();
		dfs(root);

		return ans;
    }
};

第十八题：

Binary Tree Postorder Traversal

地址：http://oj.leetcode.com/problems/binary-tree-postorder-traversal/

分析：PostOrder

代码：

(C++版)

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
vector<int> ans;
class Solution 
{
public:
	static void dfs(TreeNode *root)
	{
		if (root == NULL) return;
		dfs(root->left);
		dfs(root->right);
		ans.push_back(root->val);
	}
    vector<int> postorderTraversal(TreeNode *root) 
	{
		ans.clear();
        dfs(root);

		return ans;
    }
};

第十九题

Surrounded Regions

地址：http://oj.leetcode.com/problems/surrounded-regions/

分析：反向思考，只BFS周边的'O

代码：

(C++版)

struct Point_
{
	int x, y;
	Point_(int a,int b):x(a), y(b){}
};

class Solution 
{
public:
    void solve(vector<vector<char>> &board) 
	{
		int size1, size2;
		int i, j;

		size1 = board.size();
		if (size1 == 0) return ;
		size2 = board[0].size();
		vector<Point_> vec;
		for (i = 0; i < size1; i++)
		{
			for (j = 0; j < size2; j++)
			{
				if (board[i][j] == 'O' && (i == 0 || i == size1-1 || j == 0 || j == size2-1))
				{
					board[i][j] = 'D';
					vec.push_back(Point_(i,j));
				}
			}
		}
		const int dir[4][2] = {1,0,-1,0,0,1,0,-1};
		while (!vec.empty())
		{
			Point_ p = vec.back();
			vec.pop_back();
			for (i = 0; i < 4; i++)
			{
				int xx = p.x+dir[i][0];
				int yy = p.y+dir[i][1];
				if (xx < 0 || xx >= size1 || yy < 0 || yy >= size2 || board[xx][yy] != 'O') continue;
				vec.push_back(Point_(xx, yy));
				board[xx][yy] = 'D';
			}
		}
		for (i = 0; i < size1; i++)
			for (j = 0; j < size2; j++)
				board[i][j] = board[i][j] == 'D' ? 'O' : 'X';
		#if(0)
		for (i = 0; i < size1; i++)
		{
			for (j = 0; j < size2; j++)
				putchar(board[i][j]);
			puts("");
		}
		#endif
	}
};

第二十题

Word Search

地址：http://oj.leetcode.com/problems/word-search/

分析：DFS，不过感觉OJ有点坑。

代码：

(C++版)

int visit[110][110];
string target;
int col, row;
int len;
const int dir[4][2] = {1,0,-1,0,0,1,0,-1};
int dfs(vector<vector<char> > &board, int x,int y, int k)
{
	int i;
	
	if (board[x][y] != target[k-1]) return 0;	
	if (k == len) return 1;
	visit[x][y] = 1;
	for (i = 0; i < 4; i++)
	{
		int xx = x + dir[i][0];
		int yy = y + dir[i][1];
		if (xx < 0 || xx >= col) continue;
		if (yy < 0 || yy >= row) continue;
		if (visit[xx][yy]) continue;
		if (dfs(board, xx, yy, k+1)) return 1;
	}
	visit[x][y] = 0;

	return 0;
}

class Solution 
{
public:
    bool exist(vector<vector<char> > &board, string word) 
	{
		int i, j;
		len = word.length();
		if (word == "") return false;
		col = board.size();
		if (col == 0) return false;
		row = board[0].size();
		target = word;
		memset(visit, 0, sizeof(visit));
		for (i = 0; i < col; i++)
		{
			for (j = 0; j < row; j++)
			{
				if (dfs(board, i, j, 1)) return true;
			}
		}
		return false;
    }
};