1. 題目

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.

2. 思路

從右下往左上移動(dòng)，每次計(jì)算當(dāng)前點(diǎn)到右下的最短。

f[i, j] = grid[i, j] + min(f[i+1,j], f[i,j+1])

下邊和右邊兩條邊線(xiàn)進(jìn)行特殊處理。

3. 代碼

class Solution {

public:

// 同63題的遞歸或迭代方案, 避免超時(shí)直接采用迭代實(shí)現(xiàn)

int minPathSum(vector>& grid) {

int m = grid.size();

if (m == 0) { return 0; }

int n = grid[0].size();

if (n == 0) { return 0; }

int sol[m+1][n+1];

sol[m-1][n] = 0;

sol[m][n-1] = 0;

int int_max = numeric_limits::max();

for (int i = m-2; i >= 0; i--) {

sol[i][n] = int_max;

}

for (int j = 0; j < n - 1; j++) {

sol[m][j] = int_max;

}

for (int i = m-1; i >= 0; i--) {

for (int j = n-1; j >= 0; j--) {

sol[i][j] = grid[i][j] + min(sol[i+1][j], sol[i][j+1]);

}

}

return sol[0][0];

}

};

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