[Leetcode] #64 Minimum Path Sum

Discription:

Given a m x 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.

Solution:

//动态规划  S[i][j] = min(S[i - 1][j], S[i][j - 1]) + grid[i][j]
//我的 直接在原二维数组上进行修改
int minPathSum(vector<vector<int>>& grid) {
if (grid.empty()) return 0;
int rows = grid.size();
int cols = grid[0].size();
for (int i = 1; i < rows; i++)
grid[i][0] += grid[i - 1][0];
for (int i = 1; i < cols; i++)
grid[0][i] += grid[0][i - 1];

for (int i = 1; i < rows; i++){
for (int j = 1; j < cols; j++){
grid[i][j] += min(grid[i - 1][j], grid[i][j - 1]);
}
}
return grid[rows - 1][cols - 1];
}

//新建一个二维数组存数据
int minPathSum(vector<vector<int>>& grid) {
int m = grid.size();
int n = grid[0].size();
vector<vector<int> > sum(m, vector<int>(n, grid[0][0]));
for (int i = 1; i < m; i++)
sum[i][0] = sum[i - 1][0] + grid[i][0];
for (int j = 1; j < n; j++)
sum[0][j] = sum[0][j - 1] + grid[0][j];
for (int i = 1; i < m; i++)
for (int j = 1; j < n; j++)
sum[i][j] = min(sum[i - 1][j], sum[i][j - 1]) + grid[i][j];
return sum[m - 1][n - 1];
}

//只用一个行向量存取数据  空间复杂度降低
int minPathSum(vector<vector<int>>& grid) {
int m = grid.size();
int n = grid[0].size();
vector<int> cur(m, grid[0][0]);
for (int i = 1; i < m; i++)
cur[i] = cur[i - 1] + grid[i][0];
for (int j = 1; j < n; j++) {
cur[0] += grid[0][j];
for (int i = 1; i < m; i++)
cur[i] = min(cur[i - 1], cur[i]) + grid[i][j];
}
return cur[m - 1];
}

GitHub-Leetcode:https://github.com/wenwu313/LeetCode
参考：https://leetcode.com/problems/minimum-path-sum/?tab=Solutions