题目描述：

There is a brick wall in front of you. The wall is rectangular and has several rows of bricks. The bricks have the same height but different width. You want to draw a vertical line from the top to the bottom and cross the least bricks.

The brick wall is represented by a list of rows. Each row is a list of integers representing the width of each brick in this row from left to right.

If your line go through the edge of a brick, then the brick is not considered as crossed. You need to find out how to draw the line to cross the least bricks and return the number of crossed bricks.

You cannot draw a line just along one of the two vertical edges of the wall, in which case the line will obviously cross no bricks.

Example:

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Input: 
[[1,2,2,1],
[3,1,2],
[1,3,2],
[2,4],
[3,1,2],
[1,3,1,1]]
Output: 2
Explanation:

Note:

  1. The width sum of bricks in different rows are the same and won’t exceed INT_MAX.
  2. The number of bricks in each row is in range [1,10,000]. The height of wall is in range [1,10,000]. Total number of bricks of the wall won’t exceed 20,000.

虽然行数与每一行的砖块数都可能达到10000，但是总的砖块数最大只有20000，又因为总的砖块间缝隙的数量一定小于砖块总数，所以可以用Hash表来记录每一个出现缝隙的位置一共出现了几次缝隙，出现缝隙次数最多的位置就是穿过砖块数最少的位置。

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class Solution {
public:
    int leastBricks(vector<vector<int>>& wall) {
        if (wall.empty()) return 0;
        int row = wall.size();
        unordered_map<int, int> occurTimes;
        int width;
        for (int i = 0; i < row; i++) {
            int sum = 0;
            for (auto j : wall[i]) {
                sum += j;
                occurTimes[sum]++;
            }
            width = sum;
        }
        occurTimes[width] = 0;
        int maxOccur = 0;
        for (auto &i : occurTimes) {
            maxOccur = max(maxOccur, i.second);
        }
        return row - maxOccur;
    }
};