Forked from Helenyin123/84. Largest Rectangle in Histogram(#).java
Created
October 23, 2017 00:46
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| public class Solution { | |
| public int largestRectangleArea(int[] heights) { | |
| if (heights == null || heights.length < 1) return 0; | |
| int res = 0; | |
| int len = heights.length; | |
| Stack<Integer> stack = new Stack<>(); | |
| stack.push(-1); | |
| for (int i = 0; i < len; i++) { | |
| while (stack.peek() != -1 && heights[i] <= heights[stack.peek()]) { | |
| res = Math.max(res, heights[stack.pop()] * (i - stack.peek() - 1)); | |
| } | |
| stack.push(i); | |
| } | |
| while (stack.peek() != -1) { | |
| res = Math.max(res, heights[stack.pop()] * (len - stack.peek() - 1)); | |
| } | |
| return res; | |
| } | |
| } |
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| //divide and conquer | |
| public class Solution { | |
| //int max=0; | |
| public int largestRectangleArea(int[] heights) { | |
| if(heights==null|heights.length==0) return 0; | |
| return divide(heights,0,heights.length-1); | |
| //return max; | |
| } | |
| public int divide(int[] heights,int start,int end){ | |
| if(start>end) return 0; | |
| if(start==end) return heights[start]; | |
| boolean sorted=true; | |
| int min=start; | |
| //int max=0; | |
| for(int i=start+1;i<=end;i++){ | |
| if(heights[i]<heights[i-1]) sorted=false; | |
| if(heights[i]<heights[min]) min=i; | |
| } | |
| if(sorted){ | |
| int max=0; | |
| for(int i=start;i<=end;i++){ | |
| max=Math.max(max,heights[i]*(end-i+1)); | |
| } | |
| return max; | |
| } | |
| int l=divide(heights,start,min-1); | |
| int r=divide(heights,min+1,end); | |
| int res=Math.max(l,r); | |
| res=Math.max(res,heights[min]*(end-start+1)); | |
| return res; | |
| } | |
| } |
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| public class Solution { | |
| public int largestRectangleArea(int[] height) { | |
| int len = height.length; | |
| Stack<Integer> s = new Stack<Integer>(); | |
| int maxArea = 0; | |
| for(int i = 0; i <= len; i++){ | |
| int h = (i == len ? 0 : height[i]); | |
| if(s.isEmpty() || h >= height[s.peek()]){ | |
| s.push(i); | |
| }else{ | |
| int tp = s.pop(); | |
| maxArea = Math.max(maxArea, height[tp] * (s.isEmpty() ? i : i - 1 - s.peek())); | |
| i--; | |
| } | |
| } | |
| return maxArea; | |
| } | |
| } |
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