[LeetCode] Largest Rectangle in Histogram
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.
辅助栈：方法1（扫描3遍）
1 class Solution { 2 public: 3 int largestRectangleArea(vector<int> &height) { 4 // Start typing your C/C++ solution below 5 // DO NOT write int main() function 6 vector<int> area(height.size()); 7 8 stack<int> s; 9 for(int i = 0; i < height.size(); i++) 10 if (s.empty()) 11 { 12 s.push(i); 13 area[i] = 0; 14 } 15 else 16 { 17 while(!s.empty()) 18 { 19 if (height[i] <= height[s.top()]) 20 { 21 s.pop(); 22 } 23 else 24 break; 25 } 26 27 if (s.empty()) 28 area[i] = i; 29 else 30 area[i] = i - s.top() - 1; 31 s.push(i); 32 } 33 34 while(!s.empty()) 35 s.pop(); 36 37 for(int i = height.size() - 1; i >= 0; i--) 38 if (s.empty()) 39 { 40 s.push(i); 41 area[i] += 0; 42 } 43 else 44 { 45 while(!s.empty()) 46 { 47 if (height[i] <= height[s.top()]) 48 { 49 s.pop(); 50 } 51 else 52 break; 53 } 54 55 if (s.empty()) 56 area[i] += height.size() - i - 1; 57 else 58 area[i] += s.top() - i - 1; 59 s.push(i); 60 } 61 62 int maxArea = 0; 63 for(int i = 0; i < area.size(); i++) 64 { 65 maxArea = max(maxArea, height[i] * (area[i] + 1)); 66 } 67 68 return maxArea; 69 } 70 };
辅助栈：方法3
1 class Solution { 2 public: 3 int largestRectangleArea(vector<int> &height) { 4 // Start typing your C/C++ solution below 5 // DO NOT write int main() function 6 int maxArea = 0; 7 stack<int> s; 8 9 for(int i = 0; i < height.size(); i++) 10 if (s.empty()) 11 { 12 s.push(i); 13 } 14 else 15 { 16 while(!s.empty()) 17 { 18 if (height[s.top()] <= height[i]) 19 { 20 s.push(i); 21 break; 22 } 23 else 24 { 25 int index = s.top(); 26 s.pop(); 27 int leftWidth = s.empty() ? index : index - s.top() - 1; 28 int rightWidth = i - index - 1; 29 maxArea = max(maxArea, height[index] * (leftWidth + rightWidth + 1)); 30 } 31 } 32 33 if (s.empty()) 34 s.push(i); 35 } 36 37 int rightIndex = s.empty() ? 0 : s.top() + 1; 38 39 while(!s.empty()) 40 { 41 int index = s.top(); 42 s.pop(); 43 int leftWidth = s.empty() ? index : index - s.top() - 1; 44 int rightWidth = rightIndex - index - 1; 45 maxArea = max(maxArea, height[index] * (leftWidth + rightWidth + 1)); 46 } 47 48 49 return maxArea; 50 } 51 };
用辅助栈：方法2（扫描一遍）
1 struct Node 2 { 3 int val; 4 int index; 5 int area; 6 Node(){} 7 Node(int v, int idx):val(v), index(idx){} 8 }; 9 10 class Solution { 11 public: 12 int largestRectangleArea(vector<int> &height) { 13 // Start typing your C/C++ solution below 14 // DO NOT write int main() function 15 stack<Node> s; 16 int maxArea = 0; 17 for(int i = 0; i < height.size(); i++) 18 if (s.empty()) 19 s.push(Node(height[i], i)); 20 else 21 { 22 while(!s.empty()) 23 { 24 Node node = s.top(); 25 if (node.val <= height[i]) 26 { 27 s.push(Node(height[i], i)); 28 break; 29 } 30 else 31 { 32 s.pop(); 33 int leftIndex = s.empty() ? 0 : s.top().index + 1; 34 maxArea = max(maxArea, (i - node.index + node.index - leftIndex) * node.val); 35 } 36 } 37 38 if (s.empty()) 39 s.push(Node(height[i], i)); 40 } 41 42 int index = height.size(); 43 while(!s.empty()) 44 { 45 Node node = s.top(); 46 s.pop(); 47 int leftIndex = s.empty() ? 0 : s.top().index + 1; 48 maxArea = max(maxArea, (index - node.index + node.index - leftIndex) * node.val); 49 } 50 51 return maxArea; 52 } 53 };
