算法打卡day52|单调栈篇03| 84.柱状图中最大的矩形

class Solution {
    public int largestRectangleArea(int[] heights) {
        int length = heights.length;
        int[] minLeftIndex = new int [length];
        int[] minRightIndex = new int [length];
        
        minLeftIndex[0] = -1 ;// 记录左边第一个小于该柱子的下标
        for (int i = 1; i < length; i++) {
            int t = i - 1;
            // 这里用while来不断向右寻找
            while (t >= 0 && heights[t] >= heights[i]) t = minLeftIndex[t];
            minLeftIndex[i] = t;
        }
 
        // 记录每个柱子右边第一个小于该柱子的下标
        minRightIndex[length - 1] = length;
        for (int i = length - 2; i >= 0; i--) {
            int t = i + 1;
            while(t < length && heights[t] >= heights[i]) t = minRightIndex[t];
            minRightIndex[i] = t;
        }
 
        
        int result = 0;
        for (int i = 0; i < length; i++) {// 求和
            int sum = heights[i] * (minRightIndex[i] - minLeftIndex[i] - 1);
            result = Math.max(sum, result);
        }
        return result;
    }
}

class Solution {
    int largestRectangleArea(int[] heights) {
        Stack st = new Stack();
        
        // 数组扩容，在头和尾各加入一个元素
        int [] newHeights = new int[heights.length + 2];
        newHeights[0] = 0;
        newHeights[newHeights.length - 1] = 0;
        for (int index = 0; index < heights.length; index++){
            newHeights[index + 1] = heights[index];
        }
 
        heights = newHeights;
        
        st.push(0);
        int result = 0;
 
        // 第一个元素已经入栈，从下标1开始
        for (int i = 1; i < heights.length; i++) {
            // 注意heights[i] 是和heights[st.top()] 比较 ，st.top()是下标
            if (heights[i] > heights[st.peek()]) {
                st.push(i);
            } else if (heights[i] == heights[st.peek()]) {
                st.pop();
                st.push(i);
            } else {
                while (heights[i] < heights[st.peek()]) { //循环比较
                    int mid = st.peek();
                    st.pop();
                    int left = st.peek();
                    int right = i;
                    int w = right - left - 1;
                    int h = heights[mid];
                    result = Math.max(result, w * h);
                }
                st.push(i);
            }
        }
        return result;
    }
}