这道题如果用暴力解法的话,枚举每一个矩形,然后来计算其面积。但是会报超时错误🙅
然后参考了一下;leetcode的官方题解
class Solution {
public:
int largestRectangleArea(vector<int>& heights) {
int n = heights.size();
vector<int> left(n), right(n);
stack<int> mono_stack;
for (int i = 0; i < n; ++i) {
while (!mono_stack.empty() && heights[mono_stack.top()] >= heights[i]) {
mono_stack.pop();
}
left[i] = (mono_stack.empty() ? -1 : mono_stack.top());
mono_stack.push(i);
}
mono_stack = stack<int>();
for (int i = n - 1; i >= 0; --i) {
while (!mono_stack.empty() && heights[mono_stack.top()] >= heights[i]) {
mono_stack.pop();
}
right[i] = (mono_stack.empty() ? n : mono_stack.top());
mono_stack.push(i);
}
int ans = 0;
for (int i = 0; i < n; ++i) {
ans = max(ans, (right[i] - left[i] - 1) * heights[i]);
}
return ans;
}
};
五月为啥还有31天,坚持一个月打卡真的好不容易啊。。。而且自己的论文还要二辩了。。。😭