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 heights = [2,1,5,6,2,3],
return 10.
思路，单调栈，维护一个递增的栈。 时间复杂度O(n)
class Solution {
public:
struct node {
int x;
int left;
int right;
};
int largestRectangleArea(vector<int>& heights) {
vector<node> st;
int n = heights.size();
if (n == 0) return 0;
if (n == 1) return heights[0];
vector<int> sum(n);
for (int i = 0; i < n; ++i) {
if (i > 0) sum[i] = sum[i-1] + 1;
else sum[i] = 1;
}
int ans = 0;
for (int i = 0; i < n; ++i) {
if (st.empty() || st.back().x < heights[i]) {
st.push_back({heights[i],i,i});
} else {
int ileft = 0, iright = 0;
while (!st.empty()) {
node v = st.back();
if (v.x >= heights[i]) {
st.pop_back();
ileft = v.left;
if (st.size()) st.back().right = v.right;
int tmp = (sum[v.right]-sum[v.left-1])*v.x;
if (tmp > ans) ans = tmp;
} else break;
}
st.push_back({heights[i],ileft,i});
}
}
while (!st.empty()) {
node v = st.back();
st.pop_back();
if (st.size()) st.back().right = v.right;
int tmp = (sum[v.right]-sum[v.left-1])*v.x;
if (tmp > ans) ans = tmp;
}
return ans;
}
};
class Solution {
public:
int largestRectangleArea(vector<int>& heights) {
heights.push_back(0);
int n = heights.size();
stack<int> st;
int ans = 0;
for (int i = 0; i < n; ++i) {
while (!st.empty() && heights[i] <= heights[st.top()]) {
int top = st.top();
st.pop();
int l;
if (st.empty()) l = - 1;
else l = st.top();
ans = max(ans, (i - l - 1) * heights[top]);
}
st.push(i);
}
return ans;
}
};