Minimum Path Sum
描述
Given a m × 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
分析
跟第 ??? 节 Unique Paths 很类似。
设状态为f[i][j]，表示从起点(0,0)到达(i,j)的最小路径和，则状态转移方程为：
f[i][j]=min(f[i-1][j], f[i][j-1])+grid[i][j]
备忘录法
// Minimum Path Sum
// 备忘录法
public class Solution {
public int minPathSum(int[][] grid) {
final int m = grid.length;
final int n = grid[0].length;
this.f = new int[m][n];
for (int i = 0; i < m; ++i) Arrays.fill(f[i], -1);
return dfs(grid, m-1, n-1);
}
private int dfs(int[][] grid, int x, int y) {
if (x < 0 || y < 0) return Integer.MAX_VALUE; // 越界，终止条件，注意，不是0
if (x == 0 && y == 0) return grid[0][0]; // 回到起点，收敛条件
return Math.min(getOrUpdate(grid, x - 1, y),
getOrUpdate(grid, x, y - 1)) + grid[x][y];
}
private int getOrUpdate(int[][] grid, int x, int y) {
if (x < 0 || y < 0) return Integer.MAX_VALUE; // 越界，注意，不是0
if (f[x][y] >= 0) return f[x][y];
else return f[x][y] = dfs(grid, x, y);
}
private int[][] f; // 缓存
}动规
// Minimum Path Sum
// 二维动规
public class Solution {
public int minPathSum(int[][] grid) {
final int m = grid.length;
final int n = grid[0].length;
if (m == 0) return 0;
int[][] f = new int[m][n];
f[0][0] = grid[0][0];
for (int i = 1; i < m; i++) {
f[i][0] = f[i - 1][0] + grid[i][0];
}
for (int i = 1; i < n; i++) {
f[0][i] = f[0][i - 1] + grid[0][i];
}
for (int i = 1; i < m; i++) {
for (int j = 1; j < n; j++) {
f[i][j] = Math.min(f[i - 1][j], f[i][j - 1]) + grid[i][j];
}
}
return f[m - 1][n - 1];
}
}动规+滚动数组
// Minimum Path Sum
// 二维动规+滚动数组
public class Solution {
public int minPathSum(int[][] grid) {
final int m = grid.length;
final int n = grid[0].length;
int[] f = new int[n];
Arrays.fill(f, Integer.MAX_VALUE); // 初始值是 INT_MAX，因为后面用了min函数。
f[0] = 0;
for (int i = 0; i < m; i++) {
f[0] += grid[i][0];
for (int j = 1; j < n; j++) {
// 左边的f[j]，表示更新后的f[j]，与公式中的f[i[[j]对应
// 右边的f[j]，表示老的f[j]，与公式中的f[i-1][j]对应
f[j] = Math.min(f[j - 1], f[j]) + grid[i][j];
}
}
return f[n - 1];
}
}