Last Updated on 2022-12-13 by Clay
題目
Given an n x n array of integers matrix, return the minimum sum of any falling path through matrix.
A falling path starts at any element in the first row and chooses the element in the next row that is either directly below or diagonally left/right. Specifically, the next element from position (row, col) will be (row + 1, col - 1), (row + 1, col), or (row + 1, col + 1).
Example 1:
Input: matrix = [[2,1,3],[6,5,4],[7,8,9]] Output: 13 Explanation: There are two falling paths with a minimum sum as shown.
Example 2:
Input: matrix = [[-19,57],[-40,-5]] Output: -59 Explanation: The falling path with a minimum sum is shown.
Constraints:
n == matrix.length == matrix[i].length1 <= n <= 100-100 <= matrix[i][j] <= 100
題目給定一個正方形尺寸為 n 的陣列，我們要找出一條路徑，該路徑上的值加總必須為陣列中可能路徑的最小值。
路徑的定義則是，假設我們現在位於第 i 個位置，我們往下走，只能從下一層的 i-1 和 i+1 的範圍內選一個點。通俗來講，就是能斜一格來走。
解題思路
累積前一層的最小值
我知道這題應該存在 DP 解，但我還是採用了直接從前一層選擇最小值選點的方式，直接加在當前值的方法。
這樣一來，我們一層層地累積最小值，在最後一層選擇最小的選點，即是我們要的答案。
累積前一層的最小值 複雜度
| Time Complexity | O(n^2) |
| Space Complexity | O(1) |
C++ 範例程式碼
class Solution {
public:
int minFallingPathSum(vector<vector<int>>& matrix) {
// Init
int dx[3] = {-1, 0, 1};
int n = matrix.size();
// Accumulate the minimum value of last row
for (int i=1; i<n; ++i) {
for (int j=0; j<n; ++j) {
int temp = INT_MAX;
for (int& x: dx) {
if (j+x < 0 || j+x >= n) {
continue;
}
temp = min(temp, matrix[i-1][j+x]);
}
matrix[i][j] += temp;
}
}
return *min_element(matrix.back().begin(), matrix.back().end());
}
};Python 範例程式碼
class Solution:
def minFallingPathSum(self, matrix: List[List[int]]) -> int:
# Init
N = len(matrix)
# Accumulate the minimum value of previous row
for i in range(1, N):
for j in range(N):
left = max(0, j-1)
right = min(N-1, j+1)
matrix[i][j] += min(matrix[i-1][left:right+1])
return min(matrix[-1])