LintCode:最小路径和

最小路径和

给定一个只含非负整数的m*n网格，找到一条从左上角到右下角的可以使数字和最小的路径。

LintCode:最小路径和

思路一：万能的枚举

m行，n列的矩阵，从左上角走到右下角，需要向下移动(m-1)步，向右移动(n-1)步，共（m+n-2）步，则路径总条数为C(m+n-2， m-1)步。

m,n较小时可行，较大时不可行。

思路二： 动态规划

dp[i][j] 表示从左上到达grid[i][j]的最小值。

dp[i][j] = min(dp[i-1][j], dp[i][j-1]) + a[i][j]

从上面过来为dp[i-1][j] + grid[i][j]

从左边过来为dp[i][j-1] + grid[i][j]

初值

dp[0][0] = grid[0][0]

dp[0][j>0] = dp[0][j-1] + grid[0][j]

dp[i>0][0] = dp[i-1][0] + grid[i][0]

复杂度：时间O(m*n), 空间O(m*n)

class Solution {
public:
/**
* @param grid: a list of lists of integers.
* @return: An integer, minimizes the sum of all numbers along its path
*/
int minPathSum(vector<vector<int> > &grid) {
// write your code here
int m = grid.size();
int n = grid[0].size();
vector<vector<int> > dp(m, vector<int>(n));
for (int i=0; i<m; i++){
for (int j=0; j<n; j++){
if (i==0){
if(j==0){
dp[i][j] = grid[i][j];
}
else{
dp[i][j] = dp[i][j-1] + grid[i][j];
}
}
else if(j==0){
dp[i][j] = dp[i-1][j] + grid[i][j];
}
else{
dp[i][j] = min(dp[i-1][j], dp[i][j-1]) + grid[i][j];
}
}
}
return dp[m-1][n-1];

}
};

dp空间优化

dp[i][j]只与dp[i-1][j], dp[i][j-1]有关

对每个i， 正向循环j

前的dp[j-1]是新的，dp[j]还是旧的

dp[j] = min(dp[j-1], dp[j] + a[i][j] 更新

空间复杂度O(n), 时间复杂度O(m*n)

class Solution {
public:
/**
* @param grid: a list of lists of integers.
* @return: An integer, minimizes the sum of all numbers along its path
*/
int minPathSum(vector<vector<int> > &grid) {
// write your code here
int m = grid.size();
int n = grid[0].size();
vector<int > dp(n);
for (int i=0; i<m; i++){
for (int j=0; j<n; j++){
if (i==0){
if(j==0){
dp[j] = grid[i][j];
}
else{
dp[j] = dp[j-1] + grid[i][j];
}
}
else 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];

}
};

参考资料

七月算法公开课 实战动态规划