2017. Grid Game
## 思路
如果第一個玩家選擇在 index i 往下走
p1會拿到 grid[0][:i+1] + grid[1][i:]
p2會拿到 max(grid[0][i+1:], grid[1][:i])
所以先計算row1=grid[0]的和
掃index更新row1 row2的值
## Code
```cpp
class Solution {
public:
long long gridGame(vector<vector<int>>& grid) {
int lenC = grid[0].size();
long long row1 = 0, row2 = 0;
for (int& num: grid[0]) {
row1 += num;
}
long long res = LONG_LONG_MAX;
for (int i=0; i<lenC; ++i) {
row1 -= grid[0][i];
res = min(res, max(row1, row2));
row2 += grid[1][i];
}
return res;
}
};
```