本文深入解析了LeetCode上从N皇后到跳跃游戏等经典算法题目的解决方案，涵盖了递归、动态规划及贪心算法等多种技术。通过C++实现，详细讲解了每道题目的核心思路与代码实现。
LeetCode Algorithm 0051 - 0055
文章目录
0051 - N-Queens (Hard)
Problem Link: https://leetcode.com/problems/n-queens/
Description
The n-queens puzzle is the problem of placing n queens on an n×n chessboard such that no two queens attack each other.
Given an integer n, return all distinct solutions to the n-queens puzzle.
Each solution contains a distinct board configuration of the n-queens’ placement, where 'Q' and '.' both indicate a queen and an empty space respectively.
Example:
Input: 4
Output: [
[".Q..", // Solution 1
"...Q",
"Q...",
"..Q."],
["..Q.", // Solution 2
"Q...",
"...Q",
".Q.."]
]
Explanation: There exist two distinct solutions to the 4-queens puzzle as shown above.
Solution C++
#pragma once
#include "pch.h"
// Problem: https://leetcode.com/problems/n-queens/
namespace P51NQueens
{
class Solution
{
private:
const char dot = '.';
const char queen = 'Q';
public:
vector<vector<string>> solveNQueens(int n)
{
vector<vector<string>> results;
vector<string> result = vector<string>(n, string(n, dot));
Solving(results, n, result, 0);
return results;
}
private:
void Solving(vector<vector<string>>& results, int n, vector<string>& result, int row)
{
if (row == n)
{
results.push_back(result);
return;
}
for (int col = 0; col < n; col++)
{
result[row][col] = queen;
bool flag = true;
for (int r = 0; r < row; r++)
{
// 同列
if (result[r][col] == queen)
{
flag = false;
break;
}
int diff = row - r;
int left = col - diff;
int right = col + diff;
// 左上方斜线
if (left >= 0 && result[r][left] == queen)
{
flag = false;
break;
}
// 右上方斜线
if (right < n && result[r][right] == queen)
{
flag = false;
break;
}
}
if (flag)
{
Solving(results, n, result, row + 1);
}
result[row][col] = dot;
}
}
};
}
0052 - N-Queens II (Hard)
Problem Link: https://leetcode.com/problems/n-queens-ii/
Description
The n-queens puzzle is the problem of placing n queens on an n×n chessboard such that no two queens attack each other.
Given an integer n, return all distinct solutions to the n-queens puzzle.
Example:
Input: 4
Output: 2
Explanation: There are two distinct solutions to the 4-queens puzzle as shown below.
[
[".Q..", // Solution 1
"...Q",
"Q...",
"..Q."],
["..Q.", // Solution 2
"Q...",
"...Q",
".Q.."]
]
Solution C++
#pragma once
#include "pch.h"
// Problem: https://leetcode.com/problems/n-queens-ii/
namespace P51NQueensII
{
class Solution
{
private:
const char dot = '.';
const char queen = 'Q';
public:
int totalNQueens(int n)
{
vector<string> result = vector<string>(n, string(n, dot));
return Total(0, n, result, 0);
}
private:
int Total(int total, int n, vector<string>& result, int row)
{
if (row == n)
{
return total + 1;
}
for (int col = 0; col < n; col++)
{
result[row][col] = queen;
bool flag = true;
for (int r = 0; r < row; r++)
{
// 同列
if (result[r][col] == queen)
{
flag = false;
break;
}
int diff = row - r;
int left = col - diff;
int right = col + diff;
// 左上方斜线
if (left >= 0 && result[r][left] == queen)
{
flag = false;
break;
}
// 右上方斜线
if (right < n && result[r][right] == queen)
{
flag = false;
break;
}
}
if (flag)
{
total = Total(total, n, result, row + 1);
}
result[row][col] = dot;
}
return total;
}
};
}
0053 - Maximum Subarray (Easy)
Problem Link: https://leetcode.com/problems/maximum-subarray/
Description
Given an integer array nums, find the contiguous subarray (containing at least one number) which has the largest sum and return its sum.
Example:
Input: [-2,1,-3,4,-1,2,1,-5,4],
Output: 6
Explanation: [4,-1,2,1] has the largest sum = 6.
Follow up:
If you have figured out the O(n) solution, try coding another solution using the divide and conquer approach, which is more subtle.
Solution C++
#pragma once
#include "pch.h"
// Problem: https://leetcode.com/problems/maximum-subarray/
namespace P53MaximumSubarray
{
class Solution
{
public:
int maxSubArray(vector<int>& nums)
{
#if false // 穷举 exhaustion
int maxVal = INT_MIN;
for (int i = 0; i < nums.size(); i++)
{
int sum = 0;
for (int j = i; j < nums.size(); j++)
{
sum += nums[j];
maxVal = max(sum, maxVal);
}
}
return maxVal;
#endif
#if true // O(n)
int curVal = 0;
int maxVal = INT_MIN;
for (int i = 0; i < nums.size(); i++)
{
curVal = max(nums[i], curVal + nums[i]);
maxVal = max(maxVal, curVal);
}
return maxVal;
#endif
}
};
}
0054 - Spiral Matrix (Medium)
Problem Link: https://leetcode.com/problems/spiral-matrix/
Description
Given a matrix of m x n elements ( m rows, n columns), return all elements of the matrix in spiral order.
Example 1:
Input:
[
[ 1, 2, 3 ],
[ 4, 5, 6 ],
[ 7, 8, 9 ]
]
Output: [1,2,3,6,9,8,7,4,5]
Example 2:
Input:
[
[1, 2, 3, 4],
[5, 6, 7, 8],
[9,10,11,12]
]
Output: [1,2,3,4,8,12,11,10,9,5,6,7]
Solution C++
#pragma once
#include "pch.h"
// Problem: https://leetcode.com/problems/spiral-matrix/
namespace P54SpiralMatrix
{
class Solution
{
private:
enum Direction
{
Right,
Down,
Left,
Up,
};
public:
vector<int> spiralOrder(vector<vector<int>>& matrix)
{
vector<int> result;
if (matrix.size() == 0 || matrix[0].size() == 0)
{
return result;
}
int size = matrix.size() * matrix[0].size();
Direction dir = Direction::Right;
int curRow = 0;
int startRow = 0;
int endRow = matrix.size() - 1;
int curCol = 0;
int startCol = 0;
int endCol = matrix[0].size() - 1;
#if true // 每一圈为一个循环 O(n)
while (startRow <= endRow && startCol <= endCol)
{
// Direction::Right
for (curCol = startCol; curCol <= endCol; curCol++)
{
result.push_back(matrix[startRow][curCol]);
}
// Direction::Down
for (curRow = startRow + 1; curRow <= endRow; curRow++)
{
result.push_back(matrix[curRow][endCol]);
}
if (startRow < endRow && startCol < endCol)
{
// Direction::Left
for (curCol = endCol - 1; curCol >= startCol; curCol--)
{
result.push_back(matrix[endRow][curCol]);
}
// Direction::Up
for (curRow = endRow - 1; curRow > startRow; curRow--)
{
result.push_back(matrix[curRow][startCol]);
}
}
startCol++;
endCol--;
startRow++;
endRow--;
}
#endif
#if false // 每个节点为一个循环 O(n)
for (int i = 0; i < size; i++)
{
result.push_back(matrix[curRow][curCol]);
switch (dir)
{
case Direction::Right:
if (curCol == endCol)
{
dir = Direction::Down;
endCol--;
curRow++;
}
else
{
curCol++;
}
break;
case Direction::Down:
if (curRow == endRow)
{
dir = Direction::Left;
endRow--;
curCol--;
}
else
{
curRow++;
}
break;
case Direction::Left:
if (curCol == startCol)
{
dir = Direction::Up;
startCol++;
curRow--;
}
else
{
curCol--;
}
break;
case Direction::Up:
if (curRow == startRow + 1)
{
dir = Direction::Right;
startRow++;
curCol++;
}
else
{
curRow--;
}
break;
}
}
#endif
return result;
}
};
}
0055 - Jump Game (Medium)
Problem Link: https://leetcode.com/problems/jump-game/
Description
Given an array of non-negative integers, you are initially positioned at the first index of the array.
Each element in the array represents your maximum jump length at that position.
Determine if you are able to reach the last index.
Example 1:
Input: [2,3,1,1,4]
Output: true
Explanation: Jump 1 step from index 0 to 1, then 3 steps to the last index.
Example 2:
Input: [3,2,1,0,4]
Output: false
Explanation: You will always arrive at index 3 no matter what. Its maximum
jump length is 0, which makes it impossible to reach the last index.
Solution C++
#pragma once
#include "pch.h"
// Problem: https://leetcode.com/problems/jump-game/
namespace P55JumpGame
{
class Solution
{
public:
bool canJump(vector<int>& nums)
{
if (nums.empty() || nums.size() == 1)
{
return true;
}
int curIndex = 0;
int largest = 0;
for (int i = 0; i < nums.size() - 1; i++)
{
largest = max(largest, i + nums[i]);
if (i == curIndex)
{
curIndex = largest;
}
if (curIndex >= nums.size() - 1)
{
return true;
}
}
return false;
}
};
}
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