64.Minimum Path Sum---dp
题目链接:https://leetcode.com/problems/minimum-path-sum/description/
题目大意:从左上到右下的路径中,找出路径和最小的路径(与62,63题相联系)。
法一:dfs,果然超时,无剪枝。代码如下:
public int minPathSum(int[][] grid) {
boolean vis[][] = new boolean[grid.length][grid[0].length];
int f[][] = {{1, 0}, {0, 1}};
return dfs(grid, 0, 0, grid[0][0], Integer.MAX_VALUE, vis, f);
}
public static int dfs(int[][] grid, int x, int y, int sum, int res, boolean vis[][], int f[][]) {
if(sum >= res) {
return res;
}
if(x == grid.length - 1 && y == grid[0].length - 1) {
if(sum < res) {
res = sum;
}
return res;
}
for(int i = 0; i < 2; i++) {
int cnt_x = x + f[i][0];
int cnt_y = y + f[i][1];
if(cnt_x < grid.length && cnt_y < grid[0].length && vis[cnt_x][cnt_y] == false) {
vis[cnt_x][cnt_y] = true;
res = dfs(grid, cnt_x, cnt_y, sum + grid[cnt_x][cnt_y], res, vis, f);
vis[cnt_x][cnt_y] = false;
}
}
return res;
}
法二:dp,模仿62的二维dp,只是这里dp[i][j]表示到终点坐标为[i,j]的最短路径和,dp公式为dp[i][j] = min(dp[i-1][j],dp[i][j-1]) + grid[i][j]。代码如下(耗时9ms):
public int minPathSum(int[][] grid) {
int dp[][] = new int[grid.length][grid[0].length];
//初始化第一列
dp[0][0] = dp[0][0] = grid[0][0];
for(int i = 1; i < grid.length; i++) {
dp[i][0] = dp[i - 1][0] + grid[i][0];
}
//初始化第一行
for(int i = 1; i < grid[0].length; i++) {
dp[0][i] = dp[0][i - 1] + grid[0][i];
}
//计算dp
for(int i = 1; i < grid.length; i++) {
for(int j = 1; j < grid[0].length; j++) {
//取从左边和从上边到达的最短路径+当前值
dp[i][j] = Math.min(dp[i - 1][j], dp[i][j - 1]) + grid[i][j];
}
}
return dp[grid.length - 1][grid[0].length - 1];
}
法三:一维dp,代码如下(耗时9ms):
public int minPathSum(int[][] grid) {
int[] dp = new int[grid[0].length];
//初始化第一行
dp[0] = grid[0][0];
for(int j = 1; j < grid[0].length; j++) {
dp[j] = grid[0][j] + dp[j - 1];
}
//从第一行第0列开始计算
for(int i = 1; i < grid.length; i++) {
//计算第0列
dp[0] += grid[i][0];
//从第1列开始
for(int j = 1; j < grid[0].length; j++) {
dp[j] = Math.min(dp[j - 1], dp[j]) + grid[i][j];
}
}
return dp[grid[0].length - 1];
}
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