Re: [閒聊] 每日leetcode

[sixB (6B)]

a十個零叔叔說刷題沒救了

可是我也不知道我還能幹嘛QQ

要轉什麼方向發展比較好啊

真的要學餐飲了嗎

703.

今天這題就pq

沒想清楚倒過來放就好

想說要random access

所以用vector做二分搜插入

不放扣出來醜人惹：（

1568.

也把昨天那題寫完了

突然就順順的跑完

縫縫補補分3個case

不過solution好像也差不多

昨天一直沒想清楚又一直改

以為每格都掃一次會很慢

constraint其實也才30*30

簡單來說就是我太笨

class Solution {
public:
    int minDays(vector<vector<int>>& grid) {
        int n = grid.size(), m = grid[0].size();
        bool flag = false;
        vector<vector<int>> check0 = grid;
        int land = 0;
        for(int i = 0; i < n; i++){
            for(int j = 0; j < m; j++){
                // 1 cut
                if(grid[i][j] == 1) {
                    cut(check0, grid, i, j);
                    flag = true;
                    break;
                }
            }
            if(flag) break;
        }
        for(int i = 0; i < n; i++){
            for(int j = 0; j < m; j++){
                if(check0[i][j] == 1) {
                    return 0;
                }
                if(grid[i][j] == 1) land++;
            }
        }
        if(land <= 2) return land;

        for(int i = 0; i < n; i++){
            for(int j = 0; j < m; j++){
                // 1 cut
                vector<vector<int>> check = grid;
                if(grid[i][j] == 1){
                    grid[i][j] = 0;
                    check[i][j] = 0;
                    if(i < n-1 and grid[i+1][j] == 1)
                        cut(check, grid, i+1, j);
                    else if(i > 0 and grid[i-1][j] == 1)
                        cut(check, grid, i-1, j);
                    else if(j < m-1 and grid[i][j+1] == 1)
                        cut(check, grid, i, j+1);
                    else if(j > 0 and grid[i][j-1] == 1)
                        cut(check, grid, i, j-1);
                    grid[i][j] = 1;

                    if((i < n-1 and check[i+1][j]) or\
                     (i > 0 and check[i-1][j]) or\
                     (j < m-1 and check[i][j+1]) or\
                     (j > 0 and check[i][j-1])){
                        cout << "COOL\n";
                        cout << i << " " << j << '\n';
                        return 1;
                    }
                }
            }
        }
        return 2;
    }

    void cut(vector<vector<int>>& check, vector<vector<int>>& grid, int i, int j){
        if(i < 0 or i >= grid.size() or j < 0 or j >= grid[0].size()) return;
        if(check[i][j] == 0) return;
        check[i][j] = 0;
        cut(check, grid, i-1, j);
        cut(check, grid, i+1, j);
        cut(check, grid, i, j-1);
        cut(check, grid, i, j+1);
    }
};

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→ sixB: 上下左右可以用陣列
→ sixB: 多寫一個迴圈好像也沒少幾行
噓 a000000000: 尼可以去豬屎屋寫firmware還啥的
→ a000000000: 不過填reg這種事跟刷題關C沒很大
→ a000000000: 我最近還看過一堆矽谷刷題仔淪落到去豬屎屋作P&R
→ a000000000: 比CAD還慘  CAD好歹算寫程式der
推 ttucse: CAD是啥，寫verilog那種的嗎？
→ a000000000: 就在豬屎屋管各種tool的  像cadence synopsys賣得tool
推 ttucse: 噢噢，所以可能是寫shell script或python嗎？
噓 a000000000: 都有R
→ ttucse: 瞭了，感謝。
→ devilkool: 刷題仔會p&r喔@@
噓 a000000000: 尼掛EE學位摸點vlsi也很正常八
→ sixB: 好像真的只能寫韌體惹 嵌入式還有機會ㄇ