版权声明:自由转载-非商用-非衍生-保持署名 https://blog.csdn.net/mgsweet/article/details/78731739
解题思路
一开始没看只能从上到下从左到右导致花费了不少时间编写了一堆垃圾,有着
You can only move either down or right at any point in time.
这个条件,整个思路就明晰了。利用动态规划的方法,首先研究子问题得出:
设v[i][j]为到达grid[i][j]的最短路径,则v[i][j] = min{v[i][j - 1], v[i - 1][j]} + grid[i][j],这样我们只要把第一行第一列初始化再执行算法就可以了。
代码
class Solution {
public:
int minPathSum(vector<vector<int>>& grid) {
vector<vector<int>> v(grid.size());
for (int i = 0; i < grid.size(); i++) {
v[i].resize(grid[0].size());
}
v[0][0] = grid[0][0];
for (int i = 1; i < grid.size(); i++) {
v[i][0] = v[i - 1][0] + grid[i][0];
}
for (int i = 1; i < grid[0].size(); i++) {
v[0][i] = v[0][i - 1] + grid[0][i];
}
for (int i = 1; i < grid.size(); i++) {
for (int j = 1; j < grid[0].size(); j++) {
v[i][j] = min(v[i - 1][j], v[i][j - 1]) + grid[i][j];
}
}
return v[grid.size() - 1][grid[0].size() - 1];
}
};