目录
【力扣】84. 柱状图中最大的矩形
给定 n 个非负整数，用来表示柱状图中各个柱子的高度。每个柱子彼此相邻，且宽度为 1 。求在该柱状图中，能够勾勒出来的矩形的最大面积。
示例 1:
输入：heights = [2,1,5,6,2,3]
输出：10
解释：最大的矩形为图中红色区域，面积为 10
示例 2：
输入： heights = [2,4]
输出： 4
提示：
1 <= heights.length <= 1 0 5 10^5 105
0 <= heights[i] <= 1 0 4 10^4 104
题解
暴力求解
java
public static int largestRectangleArea(int[] heights) {
int sum = 0;
for (int i = 0; i < heights.length; i++) {
int left = i;
int right = i;
//找当前遍历元素之前第一个比它小的元素
while (left >= 0) {
if (heights[left] < heights[i]) {
break;
}
left--;
}
//找当前遍历元素之后第一个比它小的元素
while (right < heights.length) {
if (heights[right] < heights[i]) {
break;
}
right++;
}
int w = right - left - 1;
int h = heights[i];
sum = Math.max(sum, w * h);
}
return sum;
}双指针
java
public class Solution {
public static int largestRectangleArea(int[] heights) {
int[] minLeftIndex = new int[heights.length];
int[] minRightIndex = new int[heights.length];
// 记录左边第一个小于该柱子的下标
minLeftIndex[0] = -1;
for (int i = 1; i < heights.length; i++) {
int t = i - 1;
// 这里不是用if，而是不断向右寻找的过程
while (t >= 0 && heights[t] >= heights[i]) {
t = minLeftIndex[t];
}
minLeftIndex[i] = t;
}
// 记录每个柱子右边第一个小于该柱子的下标
minRightIndex[heights.length - 1] = heights.length;
for (int i = heights.length - 2; i >= 0; i--) {
int t = i + 1;
while (t < heights.length && heights[t] >= heights[i]) {
t = minRightIndex[t];
}
minRightIndex[i] = t;
}
/*for (int a : minLeftIndex) {
System.out.println(a);
}
System.out.println("______________________________");
for (int a : minRightIndex) {
System.out.println(a);
}*/
// 求和
int result = 0;
for (int i = 0; i < heights.length; i++) {
int sum = heights[i] * (minRightIndex[i] - minLeftIndex[i] - 1);
result = Math.max(sum, result);
}
return result;
}
public static void main(String[] args) {
int[] heights = {2, 4, 2};
System.out.println(largestRectangleArea(heights));
}
}单调栈
注意：单调栈是递减的
java
class Solution {
int largestRectangleArea(int[] heights) {
Stack<Integer> st = new Stack<Integer>();
// 数组扩容，在头和尾各加入一个元素，防止只递增或者只递减的数组
int [] newHeights = new int[heights.length + 2];
newHeights[0] = 0;
newHeights[newHeights.length - 1] = 0;
for (int index = 0; index < heights.length; index++){
newHeights[index + 1] = heights[index];
}
heights = newHeights;
st.push(0);
int result = 0;
// 第一个元素已经入栈，从下标1开始
for (int i = 1; i < heights.length; i++) {
// 注意heights[i] 是和heights[st.top()] 比较 ，st.top()是下标
if (heights[i] > heights[st.peek()]) {
st.push(i);
} else if (heights[i] == heights[st.peek()]) {
st.pop(); // 这个可以加，可以不加，效果一样，思路不同
st.push(i);
} else {
while (heights[i] < heights[st.peek()]) { // 注意是while
int mid = st.peek();
st.pop();
int left = st.peek();
int right = i;
int w = right - left - 1;
int h = heights[mid];
result = Math.max(result, w * h);
}
st.push(i);
}
}
return result;
}
}