柱状图中最大的矩形

解题思路：分冶法

class Solution {
    public int largestRectangleArea(int[] heights) {
        int len = heights.length;
        if(len==0){
            return 0;
        }
        int min = minArray(heights, 0, len);
        int areaMax = heights[min]*len;
        return Math.max(areaMax, Math.max(largestRectangleArea2(heights,0,min),largestRectangleArea2(heights,min+1,len)));
    }
    public int largestRectangleArea2(int[] heights,int start,int end) {
        if(start==end) {
            return 0;
        }
        int min = minArray(heights, start, end);
        int areaMax = heights[min]*(end - start);
        return Math.max(areaMax, Math.max(largestRectangleArea2(heights,start,min),largestRectangleArea2(heights,min+1,end)));
    }
    public int minArray(int[] heights,int start, int end) {
        int min = heights[start];
        int k = start;
        for(int i=start+1;i<end;i++) {
            if(heights[i]<min) {
                min = heights[i];
                k = i;
            }
        }
        return k;
    }  
}