leetcode -- Largest Rectangle in Histogram TODO O(N)

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 =

[2,1,5,6,2,3]

.



The largest rectangle is shown in the shaded area, which has area =

10

unit.

For example,
Given height =

[2,1,5,6,2,3]

,
return

10

.

本题思路和Trapping Rain Water差不多，计算每个idx的left bound 和right bound（两个bound都需大于height[idx]）

时间复杂度为：O(n^2)

public int largestRectangleArea(int[] height) {
// Start typing your Java solution below
// DO NOT write main() function
int len = height.length;
int[] area = new int[len];

for(int i = 0; i < len; i++){
int h = height[i];

int m = i - 1;
for(; m >= 0; m --){
if(height[m] < h){
break;
}
}
m ++;

int n = i + 1;
for(; n < len; n ++){
if(height
< h){
break;
}
}
n --;

int width = (n - m) + 1;
area[i] = h * width;
}
int max = 0;
for(int i = 0; i < len; i++){
if(area[i] > max){
max = area[i];
}
}
return max;
}

可以过小数据，大数据挂在输入[1,1,1,1,1,1........] ，说明有很多重复计算，做了一个简单改进，当前高度与上一个相同时，直接将area[i] = area[i-1]（line 12-15）

public int largestRectangleArea(int[] height) {
// Start typing your Java solution below
// DO NOT write main() function
int len = height.length;
int[] area = new int[len];

for(int i = 0; i < len; i++){
int h = height[i];

12             if(i >= 1 && h == height[i - 1]){
13                 area[i] = area[i - 1];
14                 continue;
15             }

int m = i - 1;
for(; m >= 0; m --){
if(height[m] < h){
break;
}
}
m ++;

int n = i + 1;
for(; n < len; n ++){
if(height
< h){
break;
}
}
n --;

int width = (n - m) + 1;
area[i] = h * width;
}
int max = 0;
for(int i = 0; i < len; i++){
if(area[i] > max){
max = area[i];
}
}
return max;
}