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 leastbricks.
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:
Input:
[[1,2,2,1],
[3,1,2],
[1,3,2],
[2,4],
[3,1,2],
[1,3,1,1]]
Output: 2
Explanation:
Note:
- The width sum of bricks in different rows are the same and won’t exceed INT_MAX.
- 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.
这道题给了我们一个砖头墙壁，上面由不同的长度的砖头组成，让选个地方从上往下把墙劈开，使得被劈开的砖头个数最少，前提是不能从墙壁的两边劈，这样没有什么意义。这里使用一个 HashMap 来建立每一个断点的长度和其出现频率之间的映射，这样只要从断点频率出现最多的地方劈墙，损坏的板砖一定最少。遍历砖墙的每一层，新建一个变量 sum，然后从第一块转头遍历到倒数第二块，将当前转头长度累加到 sum 上，这样每次得到的 sum 就是断点的长度，将其在 HashMap 中的映射值自增1，并且每次都更新下最大的映射值到变量 mx，这样最终 mx 就是出现次数最多的断点值，在这里劈开，绝对损伤的转头数量最少，参见代码如下：
class Solution {
public:
int leastBricks(vector<vector<int>>& wall) {
int mx = 0, n = wall.size();
unordered_map<int, int> m;
for (auto &row : wall) {
int sum = 0, cnt = row.size();
for (int i = 0; i < cnt - 1; ++i) {
sum += row[i];
++m[sum];
mx = max(mx, m[sum]);
}
}
return n - mx;
}
};
Github 同步地址：
https://github.com/grandyang/leetcode/issues/554
参考资料：
https://leetcode.com/problems/brick-wall/
https://leetcode.com/problems/brick-wall/discuss/101738/C%2B%2B-6-lines-(hash-map)
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