【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文章来源:https://www.toymoban.com/news/detail-661359.html
题解
暴力求解
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
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:单调栈是递减的文章来源地址https://www.toymoban.com/news/detail-661359.html
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;
}
}
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