1.题目描述
931. Minimum Falling Path Sum
Medium
11311
Given a square array of integers A, we want the minimum sum of a falling path through A.
A falling path starts at any element in the first row, and chooses one element from each row. The next row's choice must be in a column that is different from the previous row's column by at most one.
Example 1:
Input: [[1,2,3],[4,5,6],[7,8,9]] Output: 12 Explanation: The possible falling paths are:
[1,4,7], [1,4,8], [1,5,7], [1,5,8], [1,5,9][2,4,7], [2,4,8], [2,5,7], [2,5,8], [2,5,9], [2,6,8], [2,6,9][3,5,7], [3,5,8], [3,5,9], [3,6,8], [3,6,9]
The falling path with the smallest sum is [1,4,7], so the answer is 12.
Note:
1 <= A.length == A[0].length <= 100-100 <= A[i][j] <= 100
题目大概意思就是说找一条最短的下落路径，所谓下落路径要满足这样的条件：从第一行任意一个元素开始，每行都要选择一个元素。下一行所选元素的列和上一行所选元素的列最多相差1.
2.解题思路
用动态规划解题。
用一个二维数组 int dp[ROW_NUM][COL_NUM]记录过程。其中dp[i][j]表示到达（i，j）这个点的最短路径长度。那么根据题意，就有 dp[i][j] = min（dp[i-1][j-1], dp[i-1][j], dp[i-1][j+1]）+（i，j）这一点本身的数值。dp数组第一行初始化为所给矩阵的第一行，然后用两个for循环完成对dp数组的全部赋值。最后找出dp数组最后一行最小的数并返回即可。
3.实现代码
class Solution {
public:
int minFallingPathSum(vector<vector<int>>& A) {
int sum = 0;
int dp[100][100];
memset(dp, 0, sizeof(dp));
int row = A.size();
int col = A.size();
for (int i=0; i<row; i++) {
dp[0][i] = A[0][i];
}
for (int cur_row = 1; cur_row<row; cur_row++) {
for (int cur_col=0; cur_col<col; cur_col++) {
if (cur_col-1 < 0) {
dp[cur_row][cur_col] = min(dp[cur_row-1][cur_col], dp[cur_row-1][cur_col+1]) + A[cur_row][cur_col];
}
else if (cur_col+1 > col-1) {
dp[cur_row][cur_col] = min(dp[cur_row-1][cur_col], dp[cur_row-1][cur_col-1]) + A[cur_row][cur_col];
}
else {
dp[cur_row][cur_col] = min(min(dp[cur_row-1][cur_col], dp[cur_row-1][cur_col-1]),dp[cur_row-1][cur_col+1] ) + A[cur_row][cur_col];
}
}
}
sum = dp[row-1][0];
for (int i=0; i<col; i++) {
sum = min(dp[row-1][i], sum);
}
return sum;
}
};