Given a m x n grid filled with non-negative numbers, find a path from top left to bottom right which minimizes the sum of all numbers along its path.
Note: You can only move either down or right at any point in time.
Example:
Input: [ [1,3,1], [1,5,1], [4,2,1] ] Output: 7 Explanation: Because the path 1→3→1→1→1 minimizes the sum.
一个二维矩阵,每一格有一个数值,求从左上走到右下的和最小的路径。
思路:和不同路径和思路一样
- 首先得到初始状态,可知第一行每个数值就等于前一处的值加上当前值,第一列即等于上一处的值加上当前值:
for(int i=1; i<m; i++) {
dp[i][0] = dp[i-1][0] + grid[i][0];
}
for(int i=1; i<n; i++) {
dp[0][i] = dp[0][i-1] + grid[0][i];
}
- 然后每一处等于上侧/左侧的最小dp值加上当前格的值。由递推式:dp[i][j] = min (dp[i-1][j] , dp[i][j-1] ) + grid[i][j]
for(int i=1; i<m; i++) {
for(int j=1; j<n; j++) {
dp[i][j] = min(dp[i-1][j], dp[i][j-1]) + grid[i][j];
}
}
最后返回dp[m-1][n-1]即可
总代码:
int minPathSum(vector<vector<int>>& grid) {
// 2D dp matrix
if(grid.empty())
return 0;
int m = grid.size();
int n = grid[0].size();
vector<vector<int>> dp (m, vector<int> (n, grid[0][0]));
for(int i=1; i<m; i++) {
dp[i][0] = dp[i-1][0] + grid[i][0];
}
for(int i=1; i<n; i++) {
dp[0][i] = dp[0][i-1] + grid[0][i];
}
for(int i=1; i<m; i++) {
for(int j=1; j<n; j++) {
dp[i][j] = min(dp[i-1][j], dp[i][j-1]) + grid[i][j];
}
}
return dp[m-1][n-1]; }
改进:用一位数组。先得到第一行的值,之后对每一行,如果是第一个值,则等与上一处路径和加当前值dp[j] = dp[j] + grid[i][j]
其余地方则为dp[j] = dp[j] + grid[i][j]。
总代码:
int minPathSum(vector<vector<int>>& grid) {
// 1D dp arr
if(grid.empty())
return 0;
int m = grid.size();
int n = grid[0].size();
vector<int> dp(n, grid[0][0]);
for(int i=1; i<n; i++) {
dp[i] = dp[i-1] + grid[0][i];
}
for(int i=1; i<m; i++) {
for(int j=0; j<n; j++) {
if(j==0) {
dp[j] = dp[j] + grid[i][j];
}
else
dp[j] = min(dp[j-1], dp[j]) + grid[i][j];
}
}
return dp[n-1];
}