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.
For example,
Given height = [2,1,5,6,2,3]
,
return 10
.
遍歷數組,每找到一個局部峯值,然後向前遍歷所有的值,算出共同的矩形面積,每次對比保留最大值,代碼如下:
// Pruning optimize
class Solution {
public:
int largestRectangleArea(vector<int> &height) {
int res = 0;
for (int i = 0; i < height.size(); ++i) {
if (i + 1 < height.size() && height[i] <= height[i + 1]) {
continue;
}
int minH = height[i];
for (int j = i; j >= 0; --j) {
minH = min(minH, height[j]);
int area = minH * (i - j + 1);
res = max(res, area);
}
}
return res;
}
};
也可用棧來維護一個高度遞增序列,每次遇到較小的高度就開始計算矩形面積,代碼如下
class Solution {
public:
int largestRectangleArea(vector<int>& heights) {
int res = 0;
stack<int> st;
heights.push_back(0);
for (int i = 0; i < heights.size(); ++i) {
while (!st.empty() && heights[st.top()] >= heights[i]) {
int cur = st.top(); st.pop();
res = max(res, heights[cur] * (st.empty() ? i : (i - st.top() - 1)));
}
st.push(i);
}
return res;
}
};