作者:
dont 2025-01-06 21:02:401769. 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;