Lintcode 110. 最小路径和
题目描述: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。
- 你在同一时间只能向下或者向右移动一步
属于坐标型动态规划里面最小值的问题:
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) {
int m = grid.size(), n = grid[0].size();
if (0 == m && n == 0) {
return 0;
}
vector<vector<int> > f(m, vector<int>(n));
//1. 起点
f[0][0] = grid[0][0];
//2. 边界情况
for (int i = 1; i < m; ++i) {
f[i][0] = f[i - 1][0] + grid[i][0];
}
for (int j = 1; j < n; ++j) {
f[0][j] = f[0][j - 1] + grid[0][j];
}
//3. 状态转移
for (int i = 1; i < m; ++i) {
for (int j = 1; j < n; ++j) {
f[i][j] = min(f[i - 1][j], f[i][j - 1]) + grid[i][j];
}
}
//4. 终点
return f[m - 1][n - 1];
}
};