leetCode 84.Largest Rectangle in Histogram (最大矩形直方图) 解题思路和方法
2017-07-24 14:38
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Given n non-negative integers representing the histogram's bar height where the width of each bar is 1, find the area of largest rectangle in the histogram.
Above is a histogram where width of each bar is 1, given height =
The largest rectangle is shown in the shaded area, which has area =
For example,
Given height =
return
思路:此题还是有一些难度的。刚開始想的双指针,前后扫描,可是每次求最小的高度的时候都须要遍历一次。效率上不是非常高。为o(n2)。
代码例如以下:
解法上过不了OJ,所以仅仅能在网上參看资料,最后找到资料例如以下,是用栈解决的。
Above is a histogram where width of each bar is 1, given height =
[2,1,5,6,2,3].
The largest rectangle is shown in the shaded area, which has area =
10unit.
For example,
Given height =
[2,1,5,6,2,3],
return
10.
思路:此题还是有一些难度的。刚開始想的双指针,前后扫描,可是每次求最小的高度的时候都须要遍历一次。效率上不是非常高。为o(n2)。
代码例如以下:
public class Solution { public int largestRectangleArea(int[] height) { int max = 0;//最大值 int i = 0;//左指针 int j = height.length - 1;//右指针 boolean isMinChange = true; //双指针扫描 while(i <= j){ int minHeight = Integer.MAX_VALUE;//范围内最小值 if(isMinChange){//最小值是否改变 isMinChange = false; //又一次得到最小值 for(int k = i ; k <= j;k++){ if(height[k] < minHeight){ minHeight = height[k]; } } } //面积 int area = (j - i + 1)*minHeight; if(max < area){ max = area; } if(i == j){//假设相等,则结束 break; } if(height[i] < height[j]){//左指针添加,直到比当前大 int k = i; while(k <= j && height[k] <= height[i]){ if(height[k] == minHeight){//推断最小值是否改变 isMinChange = true; } k++; } i = k; }else{//右指针减小,直到比当前大 int k = j; while( k >= i && height[k] <= height[j]){ if(height[k] == minHeight){//推断最小值是否改变 isMinChange = true; } k--; } j = k; } } return max; } }
解法上过不了OJ,所以仅仅能在网上參看资料,最后找到资料例如以下,是用栈解决的。
public class Solution { public int largestRectangleArea(int[] height) { if (height == null || height.length == 0) return 0; Stack<Integer> stHeight = new Stack<Integer>(); Stack<Integer> stIndex = new Stack<Integer>(); int max = 0; for(int i = 0; i < height.length; i++){ if(stHeight.isEmpty() || height[i] > stHeight.peek()){ stHeight.push(height[i]); stIndex.push(i); }else if(height[i] < stHeight.peek()){ int lastIndex = 0; while(!stHeight.isEmpty() && height[i] < stHeight.peek()){ lastIndex = stIndex.pop(); int area = stHeight.pop()*(i - lastIndex); if(max < area){ max = area; } } stHeight.push(height[i]); stIndex.push(lastIndex); } } while(!stHeight.isEmpty()){ int area = stHeight.pop()*(height.length - stIndex.pop()); max = max < area ? area:max; } return max; } }
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