Given n non-negative integers representing the histogram's bar height where the width of each bar is 1, find the area of largest rectangle in the histogram.
Above is a histogram where width of each bar is 1, given height = [2,1,5,6,2,3].
The largest rectangle is shown in the shaded area, which has area = 10 unit.
Example:
Input: [2,1,5,6,2,3] Output: 10
my code:(very inefficient)
class Solution {
public:
int largestRectangleArea(vector<int>& heights) {
int len = heights.size();
int res = 0;
for (int i = 0; i < len; ++i) {
int nums = 1;
for (int j = i+1; j < len; ++j) {
if (heights[j] >= heights[i])
nums++;
else
break;
}
for (int k = i-1; k >= 0; --k) {
if (heights[k] >= heights[i])
nums++;
else
break;
}
res = max(res, heights[i]*nums);
}
return res;
}
};
Runtime: 1664 ms, faster than 1.28% of C++ online submissions for Largest Rectangle in Histogram.
height efficient code:
class Solution {
public:
int largestRectangleArea(vector<int>& heights) {
int len = heights.size();
int res = 0;
stack<int> s;
heights.push_back(0);
for (int i = 0; i <= len; ++i) {
int h = i==len ? 0 : heights[i];
while (!s.empty() && h < heights[s.top()]) {
int height = heights[s.top()];
s.pop();
int start = s.empty() ? -1 : s.top();
int nums = i - start - 1;
res = max(res, height*nums);
}
s.push(i);
}
return res;
}
};
Runtime: 12 ms, faster than 45.96% of C++ online submissions for Largest Rectangle in Histogram.
stack中的数据（索引）升序进行排列，遇到小于前面的数开始计算，并将stack中比当前元素大的数pop。