102 lines
3.3 KiB
Java
102 lines
3.3 KiB
Java
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package stack;
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import java.util.Deque;
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import java.util.LinkedList;
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/**
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* 题目： 42. 接雨水 (trap)
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* 描述：给定 n 个非负整数表示每个宽度为 1 的柱子的高度图，计算按此排列的柱子，下雨之后能接多少雨水。
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* 示例 1：
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* 输入：height = [0,1,0,2,1,0,1,3,2,1,2,1]
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* 输出：6
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* 解释：上面是由数组 [0,1,0,2,1,0,1,3,2,1,2,1] 表示的高度图，在这种情况下，可以接 6 个单位的雨水（蓝色部分表示雨水）。
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*
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* 链接：https://leetcode.cn/problems/trapping-rain-water/
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*/
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//困难题 不会
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public class Trap {
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//暴力解法
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public int trap1(int[] height) {
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int sum = 0;
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for (int i = 0; i < height.length; i++) {
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// 第一个柱子和最后一个柱子不接雨水
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if (i == 0 || i == height.length - 1) {
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continue;
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}
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int rHeight = height[i]; // 记录右边柱子的最高高度
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int lHeight = height[i]; // 记录左边柱子的最高高度
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for (int r = i + 1; r < height.length; r++) {
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if (height[r] > rHeight) {
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rHeight = height[r];
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}
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}
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for (int l = i - 1; l >= 0; l--) {
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if (height[l] > lHeight) {
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lHeight = height[l];
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}
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}
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int h = Math.min(lHeight, rHeight) - height[i];
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if (h > 0) {
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sum += h;
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}
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}
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return sum;
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}
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//动态规划
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public int trap2(int[] height) {
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if (height.length <= 2) return 0;
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int n = height.length;
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int[] maxLeft = new int[n];
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int[] maxRight = new int[n];
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// 记录每个柱子左边柱子最大高度
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maxLeft[0] = height[0];
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for (int i = 1; i < n; i++) {
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maxLeft[i] = Math.max(height[i], maxLeft[i - 1]);
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}
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// 记录每个柱子右边柱子最大高度
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maxRight[n - 1] = height[n - 1];
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for (int i = n - 2; i >= 0; i--) {
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maxRight[i] = Math.max(height[i], maxRight[i + 1]);
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}
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// 求和
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int sum = 0;
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for (int i = 1; i < n-1; i++) {
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int count = Math.min(maxLeft[i-1], maxRight[i+1]) - height[i];
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if (count > 0) {
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sum += count;
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}
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}
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return sum;
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}
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//单调栈
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public int trap(int[] height) {
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int sum = 0;
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Deque<Integer> stack = new LinkedList<>();
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for (int i = 0; i < height.length; i++) {
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// 当栈不为空且当前柱子的高度大于栈顶柱子的高度时
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while (!stack.isEmpty() && height[i] > height[stack.peek()]) {
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int top = stack.pop();
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// 如果栈为空，则没有左边的柱子来接雨水
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if (stack.isEmpty()) break;
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// 计算当前柱子与栈顶柱子之间的距离
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int distance = i - stack.peek() - 1;
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// 计算可以接的雨水高度：两边较低的高度减去中间柱子的高度
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int boundedHeight = Math.min(height[i], height[stack.peek()]) - height[top];
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sum += distance * boundedHeight;
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}
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// 将当前索引入栈
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stack.push(i);
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}
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return sum;
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}
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}
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