常规暴力破解方法，时间复杂度O(n2)：

class Solution {
public:
    vector<int> minOperations(string boxes) {
        int n = boxes.size();
        vector<int> res;
        for(int i = 0;i < n;i++)
        {
            int distance = 0;
            for(int j = 0;j < n;j++)
            {
                if(i == j) continue;
                if('1' == boxes[j])
                    distance += abs(i-j);
            }
            res.push_back(distance);
        }
        return res;
    }
};

运行效率：

Runtime: 144 ms, faster than 39.87% of C++ online submissions for Minimum Number of Operations to Move All Balls to Each Box.
Memory Usage: 9.3 MB, less than 39.10% of C++ online submissions for Minimum Number of Operations to Move All Balls to Each Box.

绞尽脑汁也没有想到优化的方法，看了讨论区大神的解法：

vector<int> minOperations(string boxes) {
    vector<int> res(boxes.length()); 
    for (int i = 0, ops = 0, cnt = 0; i < boxes.length(); ++i) {
       res[i] += ops;
       cnt += boxes[i] == '1' ? 1 : 0;
       ops += cnt;
    }
    for (int i = boxes.length() - 1, ops = 0, cnt = 0; i >= 0; --i) {
        res[i] += ops;
        cnt += boxes[i] == '1' ? 1 : 0;
        ops += cnt;
    }
    return res;
}

运行效率:

Runtime: 4 ms, faster than 96.14% of C++ online submissions for Minimum Number of Operations to Move All Balls to Each Box.
Memory Usage: 8.7 MB, less than 91.82% of C++ online submissions for Minimum Number of Operations to Move All Balls to Each Box.

相对于暴力破解方法的144ms，速度提升明显。
这个解法主要从左向右遍历一遍，然后从右向左遍历一遍。从左向右遍历时只能记录当前结点左侧结点的1全部移动到当前结点需要的次数，当从右向左遍历时可以求解当前结点右侧结点的1全部移动到当前结点需要的次数。
另一种解法：

class Solution {
    public int[] minOperations(String boxes) {
        int[] answer = new int[boxes.length()];
        int left = 0, right = 0, total = 0;//左边盒子的个数，右边盒子的个数，操作数
        //计算第一个盒子的操作数
        if (boxes.charAt(0) == '1') left ++;
        for (int i = 1 ; i < boxes.length(); i++) {
            if (boxes.charAt(i) == '1') {
                right++;
                total += i;
            }
        }
        answer[0] = total;
        //根据前一个盒子的操作数，计算下一个盒子的操作数
        for (int i = 1; i < boxes.length(); i++) {
            total = total + left - right;
            if (boxes.charAt(i) == '1') {
                left ++;
                right --;
            }
            answer[i] = total;
        }
        return answer;
    }
}

上面代码最难理解的部分是total = total + left - right;
左边有left个盒子有球，右边有right个盒子有球，当移动一个格子后左边的每个盒子距离增加1，右边每个盒子距离减少1，因此才有了total = total + left - right来计算下一个盒子游动次数。
运行效率：

Runtime: 4 ms, faster than 96.14% of C++ online submissions for Minimum Number of Operations to Move All Balls to Each Box.
Memory Usage: 8.7 MB, less than 91.82% of C++ online submissions for Minimum Number of Operations to Move All Balls to Each Box.