From 6de20998c323bc67ee3c33adb930709908dc7b77 Mon Sep 17 00:00:00 2001
From: pezy <urbancpz@gmail.com>
Date: Mon, 9 Feb 2015 12:55:54 +0800
Subject: [PATCH] added 114. Largest Rectangle in Histogram

---
 114. Largest Rectangle in Histogram/README.md | 53 +++++++++++++++++++
 114. Largest Rectangle in Histogram/TEST.cc   | 18 +++++++
 .../solution.h                                | 20 +++++++
 3 files changed, 91 insertions(+)
 create mode 100644 114. Largest Rectangle in Histogram/README.md
 create mode 100644 114. Largest Rectangle in Histogram/TEST.cc
 create mode 100644 114. Largest Rectangle in Histogram/solution.h

diff --git a/114. Largest Rectangle in Histogram/README.md b/114. Largest Rectangle in Histogram/README.md
new file mode 100644
index 0000000..db59b63
--- /dev/null
+++ b/114. Largest Rectangle in Histogram/README.md	
@@ -0,0 +1,53 @@
+
+     _           _
+    | |  _   _  | |
+    | | | |_| | | |
+    | | |*|*|*| | |
+    | |_|*|*|*| | |
+    | | |*|*|*|_| |
+    |_|_|_|_|_|_|_|
+     6 2 5 4 5 1 6
+         ^ ^ ^
+     0 1 2 3 4 5 6
+
+如图所示，很直观：`3 * 4 == 12`.首先思考，我们是如何得到最大面积 12 的？
+
+我们发现，要计算 Largest Rectangle ，关键是找**短板**，如我们挑选的 4，类似的还有哪些呢？列个表：
+
+|min_height|rectangle|area|
+|:--------:|:-------:|:--:|
+|     2    |  0 -> 4 | 2*5|
+|     4    |  2 -> 4 | 4*3|
+|     1    |  0 -> 6 | 1*6|
+
+最终很明显，area 最大的即为中间那行，结果为 12.
+
+再进一步思考：
+
+1. 短板是如何摘出来的？这个很显然，是衰减发生的时候出现的。如 6->2, 5->4, 5->1.
+2. 短板的周期如何得出？细致观察后，发现是短板之间的升降分割出的。如 2->4:(2), 2->1, 4->1:(4), 2 和 4 便是分隔符。
+
+好了，现在发现了规律，但如何准确找到所需的数据结构呢？
+
+我们的第一个目的显然是要找出短板，短板的条件是：
+
+    height[i] < height[i-1]
+
+貌似迭代一遍即可，但短板的周期分隔点如何得出呢？我们不仅要找出短板，还需要留下短板以及短板的 index, 以便寻找分隔符。
+
+看来我们需要一把筛子，筛掉不符合短板条件的，留下短板。
+
+**栈**非常符合筛子的特性，不管三七二十一，push 进去，不符合则 pop 出来。
+
+```cpp
+int max_area = 0, i = 0, size = height.size();
+for (stack<int> stk; i<size || !stk.empty(); ) { // 这个的 for 循环在二叉树的题目里经常出现（DFS）
+    if (stk.empty() || (i != size && height[stk.top()] <= height[i])) stk.push(i++); // 不是短板，则 push
+    else {
+        int tp = stk.top(); stk.pop();
+        max_area = std::max(max_area, height[tp] * (stk.empty() ? i : i-stk.top()-1));
+    }
+}
+```
+
+基本逻辑很简单，5 行结束。
\ No newline at end of file
diff --git a/114. Largest Rectangle in Histogram/TEST.cc b/114. Largest Rectangle in Histogram/TEST.cc
new file mode 100644
index 0000000..4e0cc3c
--- /dev/null
+++ b/114. Largest Rectangle in Histogram/TEST.cc	
@@ -0,0 +1,18 @@
+#define CATCH_CONFIG_MAIN
+#include "../Catch/single_include/catch.hpp"
+#include "solution.h"
+
+TEST_CASE("Largest Rectangle in Histogram", "[largestRectangleArea]")
+{
+    Solution s;
+    SECTION( "common1" )
+    {
+        std::vector<int> height{6, 2, 5, 4, 5, 1, 6};
+        REQUIRE( s.largestRectangleArea(height) == 12 );
+    }
+    SECTION( "common2" )
+    {
+        std::vector<int> height{2, 1, 5, 6, 2, 3};
+        REQUIRE( s.largestRectangleArea(height) == 10 );
+    }
+}
diff --git a/114. Largest Rectangle in Histogram/solution.h b/114. Largest Rectangle in Histogram/solution.h
new file mode 100644
index 0000000..d043f70
--- /dev/null
+++ b/114. Largest Rectangle in Histogram/solution.h	
@@ -0,0 +1,20 @@
+#include <vector>
+using std::vector;
+#include <stack>
+using std::stack;
+#include <algorithm>
+using std::max;
+
+class Solution {
+public:
+    int largestRectangleArea(vector<int> &height) {
+        int max_area = 0, i = 0, size = height.size();
+        for (stack<int> stk; i<size || !stk.empty(); )
+            if (stk.empty() || (i != size && height[stk.top()] <= height[i])) stk.push(i++);
+            else {
+                int tp = stk.top(); stk.pop();
+                max_area = max(max_area, height[tp] * (stk.empty() ? i : i-stk.top()-1));
+            }
+        return max_area;
+    }
+};