Re: [閒聊] 每日leetcode

[dont (dont)]

1769. Minimum Number of Operations to Move All Balls to Each Box

## 思路

分左右兩次計算
每次移動1個index的移動次數 會是原本次數加上球的個數

## Code

```cpp
class Solution {
public:
    vector<int> minOperations(string boxes) {
        int n = boxes.size();
        vector<int> res(n, 0);

        int curr_sum=0, curr_ball=0;
        for (int i=0; i<n; ++i) {
            res[i] = curr_sum;
            curr_ball += boxes[i] == '1';
            curr_sum += curr_ball;
        }

        curr_sum=0, curr_ball=0;
        for (int i=n-1; i>=0; --i) {
            res[i] += curr_sum;
            curr_ball += boxes[i] == '1';
            curr_sum += curr_ball;
        }

        return res;
    }
};
```

--
※ 發信站: 批踢踢實業坊(ptt.cc), 來自: 203.10.99.172 (日本)
※ 文章網址: https://www.ptt.cc/bbs/Marginalman/M.1736168563.A.4B1.html
推 sustainer123: 大師 我寫超醜
推 Meaverzt: 大師 肥肥O(n)想超久才寫出來