Largest Rectangle in Histogram

Brute Force 计算每两个index之间的rectangle O(n^2)Python TLE

class Solution: def largestRectangleArea(self, heights: List[int]) -> int: n = len(heights) res = 0 for i in range(n): minV = heights[i] for j in range(i, n): minV = min(minV, heights[j]) res = max(minV*(j-i+1), res) return res

解法一 brute force + pruning optimize

如何pruning？ 只遍历局部峰值。什么是局部峰值？下一个index高度没有现在的高，就是局部峰值了。 For example, [2,1,5,6,4,3] -> 2, 6, 4在这里都是局部峰值。为什么可以这样？ 如果下一个index高度比现在高，现在可以组成的长方形，可以包括在下一个index里。 1, 5, 6为升序，我们可以省略1, 5 1 - (2, 1)形成的长方形可以包括在(2,1,5,6)里 5 - (1, 5) 形成的长方形可以包括在(1, 5, 6)里      (2, 1, 5)形成的长方形可以包括在(2,1,5,6)里

Python

class Solution: def largestRectangleArea(self, heights: List[int]) -> int: n = len(heights) res = 0 for i in range(n): if i+1<n and heights[i+1]>=heights[i]: continue minV = heights[i] for j in range(i, -1, -1): minV = min(minV, heights[j]) res = max(res, minV*(i-j+1)) return res

C++

class Solution { public: int largestRectangleArea(vector<int>& height) { int maxV = 0; for(int i =0; i< height.size(); i++){ if(i+1 < height.size()&& height[i] <= height[i+1]) continue; int minV = height[i]; for(int j =i; j>=0; j--){ minV = min(minV, height[j]); int area = minV*(i-j+1); maxV = max(maxV, area); } } return maxV; } };

解法二 DP

基本思想是找到以每一个index高度为高的最大rectangle。遍历每一位，寻找每一位的最左和最右边界。

Python

class Solution: def largestRectangleArea(self, heights: List[int]) -> int: if not heights: return 0 n = len(heights) leftLess = [0]*n leftLess[0] = -1 for i in range(1, n): p = i-1 while p>=0 and heights[p]>=heights[i]: p = leftLess[p] leftLess[i] = p rightLess = [0]*n rightLess[n-1] = n for j in range(n-2, -1, -1): p = j+1 while p<n and heights[p]>=heights[j]: p = rightLess[p] rightLess[j] = p res = 0 for i in range(n): res = max(res, heights[i]*(rightLess[i]-leftLess[i]-1)) return res

C++

class Solution { public: int largestRectangleArea(vector<int>& heights) { if(heights.empty()) return 0; int n = heights.size(), res=0; int lessFromLeft[n] = {0}; int lessFromRight[n] = {0}; lessFromLeft[0] = -1; lessFromRight[n-1] = n; for(int i=1;i<n;i++){ int p = i-1; while(p>=0 && heights[p]>=heights[i]){ p = lessFromLeft[p]; } lessFromLeft[i] = p; } for(int i=n-2;i>=0;i--){ int p = i+1; while(p<n && heights[p]>=heights[i]){ p = lessFromRight[p]; } lessFromRight[i] = p; } for(int i=0;i<n;i++){ res = max(res, heights[i]*(lessFromRight[i]-lessFromLeft[i]-1)); } return res; } };

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解法三 Stack

方法和上面相同，只不过通过stack来实现

Python

class Solution(object): def largestRectangleArea(self, heights): """ :type heights: List[int] :rtype: int """ heights.append(0) stack = [-1] res = 0 for i in range(len(heights)): while heights[i]<heights[stack[-1]]: h = heights[stack.pop()] w = i-stack[-1]-1 res = max(res, h*w) stack.append(i) return res

C++

class Solution { public: int largestRectangleArea(vector<int> &height) { int res = 0; stack<int> st; height.push_back(0); for (int i = 0; i < height.size(); ++i) { if (st.empty() || height[st.top()] < height[i]) { st.push(i); } else { int cur = st.top(); st.pop(); res = max(res, height[cur] * (st.empty() ? i : (i - st.top() - 1))); --i; } } return res; } };

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