柱狀圖中最大的矩形

難度困難783

給定 n 個非負整數，用來表示柱狀圖中各個柱子的高度。每個柱子彼此相鄰，且寬度為 1 。

求在該柱狀圖中，能夠勾勒出來的矩形的最大面積。

以上是柱狀圖的示例，其中每個柱子的寬度為 1，給定的高度為 [2,1,5,6,2,3]。

圖中陰影部分為所能勾勒出的最大矩形面積，其面積為 10 個單位

示例:

輸入: [2,1,5,6,2,3]輸出: 10

方法一：暴力
固定高度，向兩邊找邊

public int largestRectangleArea(int[] heights) {
    int maxArea = 0;
    for (int i = 0; i < heights.length; i++) {
        int left = 0;
        int right = heights.length - 1;
        for(int j = i - 1; j >= 0; j--) {
            if (heights[j] < heights[i]) {
                left = j + 1;
                break;
            }
        }
        for(int j = i + 1; j < heights.length; j++) {
            if (heights[j] < heights[i]) {
                right = j - 1;
                break;
            }
        }
        maxArea = Math.max(maxArea, heights[i] * (right -left + 1));
    }
    return maxArea;
}
 

固定邊

public int largestRectangleArea(int[] heights) {
    int maxArea = 0;
    for (int i = 0; i < heights.length; i++) {
        int minHeight = heights[i];
        for (int j = i; j <heights.length; j++) {
            minHeight = Math.min(minHeight, heights[j]);
            maxArea = Math.max(maxArea, minHeight * (j - i + 1));
        }
    }
    return maxArea;
}
 

方法二：單調棧

public int largestRectangleArea(int[] heights) {
    int maxArea = 0;
    Stack<Integer> stack = new Stack<>();
    for (int i = 0; i < heights.length; i++) {
        while (!stack.isEmpty() && heights[stack.peek()] >= heights[i]) {
            int height = heights[stack.pop()];
            int width;
            if (stack.isEmpty()) {
                width = i;
            } else {
                width = i - stack.peek() - 1;
            }
            maxArea = Math.max(maxArea, height * width);
        }
        stack.push(i);
    }
    while (!stack.isEmpty()) {
        int height = heights[stack.pop()];
        int width;
        if (stack.isEmpty()) {
            width = heights.length;
        } else {
            width = heights.length - stack.peek() - 1;
        }
        maxArea = Math.max(maxArea, height * width);
    }
    return maxArea;
}
 

哨兵

public int largestRectangleArea(int[] heights) {
    int len = heights.length;
    int maxArea = 0;
    Stack<Integer> stack = new Stack<>();
    int[] newHeights = new int[len + 2];
    System.arraycopy(heights, 0, newHeights, 1, len);
    heights = newHeights;
    len += 2;
    stack.push(0);
    for (int i = 1; i < len; i++) {
        while (heights[i] < heights[stack.peek()]) {
            int height = heights[stack.pop()];
            int width = i - stack.peek() - 1;
            maxArea = Math.max(maxArea, height * width);
        }
        stack.push(i);
    }
    return maxArea;
}