leetcode11 & leetcode84 柱状图的最大面积 & 盛最多水的容器

leetcode11 & leetcode84 柱状图的最大面积 & 盛最多水的容器

import java.util.Stack;
public class 柱状图的最大面积84 {
	//柱状图的最大面积，单调栈（最近小于）
    public static int largestRectangleArea(int[] heights) {
        if (heights == null || heights.length == 0) return 0;
        int maxArea = 0;
        //单调栈，存储index
        Stack stack = new Stack<>();
        for (int i = 0; i < heights.length; i++) {
            while (!stack.isEmpty() && heights[i] <= heights[stack.peek()]) {
                int j = stack.pop();
                int k = stack.isEmpty() ? -1 : stack.peek();
                int curArea = (i - k - 1) * heights[j]; //i - k - 1就是出栈index所能覆盖的最长长度
                maxArea = Math.max(maxArea,curArea);
            }
            stack.push(i);
        }
        while(!stack.isEmpty()){
            int j = stack.pop();
            int k = stack.isEmpty() ? -1 : stack.peek();
            int curArea = (heights.length - k - 1) * heights[j]; //这里仍未出栈 栈顶元素是数组最后一个元素
            maxArea = Math.max(curArea,maxArea);
        }
        return maxArea;
    }
    //11盛最多水的容器 可以用单调栈做吗（最近大于最近小于，这里应该是最远大于）?
    //双指针法 权衡底长和高度之间的关系，从最大底长开始，用更长的高来权衡底长的缩短

    public static int maxArea(int[] height) {
        int max = 0;
        int l = 0;
        int r = height.length - 1;
        while (l < r) {
            max = Math.max((r - l) * Math.min(height[l], height[r]), max);
            if(height[l] < height[r]){
                l++;
            }else{
                r--;
            }
        }
        return max ;
    }


    public static void main(String[] args) {
        int[] heights = {1,8,6,2,5,4,8,3,7};
        int i = maxArea(heights);
        System.out.println(i);
    }
}