【Leetcode】【DP-二維陣列】 64. Minimum Path Sum / 最小路徑和】
阿新 • • 發佈:2018-11-10
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.
一個二維矩陣,每一格有一個數值,求從左上走到右下的和最小的路徑。
思路:和不同路徑和思路一樣
- 首先得到初始狀態,可知第一行每個數值就等於前一處的值加上當前值,第一列即等於上一處的值加上當前值:
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]; }
- 然後每一處等於上側/左側的最小dp值加上當前格的值。由遞推式: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];
}
}
最後返回dp[m-1][n-1]即可
總程式碼:
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]; }
改進:用一位陣列。先得到第一行的值,之後對每一行,如果是第一個值,則等與上一處路徑和加當前值dp[j] = dp[j] + grid[i][j]
其餘地方則為dp[j] = dp[j] + grid[i][j]。
總程式碼:
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];
}