84. Largest Rectangle in Histogram
Problem
Given an array of integers heights representing the histogram's bar height where the width of each bar is 1, return the area of the largest rectangle in the histogram.
Example
Input: heights = [2,1,5,6,2,3]
Output: 10
[1] Code (22. 03. 13)
Need to Retry -> Hard 문제는 다시 풀어보자 !
// Runtime: 148 ms
// Memory Usage: 93.1 MB
// Ref : https://leetcode.com/submissions/detail/657229072
class Solution {
public int largestRectangleArea(int[] heights) {
Stack<Node> st = new Stack<>();
int ans = -1;
if (heights.length < 1) {
return heights[0];
}
st.push(new Node(0, heights[0]));
for (int i = 1; i < heights.length; i++) {
int pos = i;
while (!st.isEmpty() && st.peek().val > heights[i]) {
Node node = st.pop();
pos = node.pos;
ans = Math.max(ans, (i - node.pos) * node.val);
}
st.push(new Node(pos, heights[i]));
}
Node rightNode = new Node(heights.length, st.peek().val);
while (!st.isEmpty()) {
Node node = st.pop();
ans = Math.max(ans, (rightNode.pos - node.pos) * node.val);
}
return ans;
}
private class Node {
public int pos;
public int val;
public Node(int pos, int val) {
this.pos = pos;
this.val = val;
}
}
}
Reference Code
Code 1
// Runtime: 75 ms
// Memory Usage: 88.8 MB
// Ref : https://leetcode.com/submissions/detail/621037929
class Solution {
public int largestRectangleArea(int[] heights) {
int len = heights.length;
Stack<Integer> s = new Stack<>();
int maxArea = 0;
for (int i = 0; i <= len; i++) {
int h = (i == len ? 0 : heights[i]);
if (s.isEmpty() || h >= heights[s.peek()]) {
s.push(i);
} else {
int tp = s.pop();
maxArea = Math.max(maxArea, heights[tp] * (s.isEmpty() ? i : i - 1 - s.peek()));
i--;
}
}
return maxArea;
}
}
- 내가 푼 풀이보다 시간/공간 복잡도가 효율적이다.
Review
-
60분 소요
기본적인 아이디어를 떠올리는 데는 크게 어렵지 않았다.
그런데 거기서 디테일 로직을 구현하는데 애를 좀 먹었다. -ㅂ-