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42. Trapping Rain Water
Given n non-negative integers representing an elevation map where the width of each bar is 1, compute how much water it is able to trap after raining.
Example:
The above elevation map is represented by array [0,1,0,2,1,0,1,3,2,1,2,1]. In this case, 6 units of rain water (blue section) are being trapped. Thanks Marcos for contributing this image!
Input: [0,1,0,2,1,0,1,3,2,1,2,1] Output: 6
解题分析
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一刷
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二刷
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三刷
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/**
* Runtime: 0 ms, faster than 100.00% of Java online submissions for Trapping Rain Water.
* Memory Usage: 38.3 MB, less than 91.10% of Java online submissions for Trapping Rain Water.
*
* Copy from: https://leetcode-cn.com/problems/trapping-rain-water/solution/xiang-xi-tong-su-de-si-lu-fen-xi-duo-jie-fa-by-w-8/[详细通俗的思路分析，多解法 - 接雨水 - 力扣（LeetCode）]
*
* @author D瓜哥 · https://www.diguage.com
* @since 2019-07-26 08:49
*/
public int trap(int[] height) {
int left = 0, right = height.length - 1;
int result = 0;
int leftMax = 0, rightMax = 0;
while (left < right) {
if (height[left] < height[right]) {
if (height[left] >= leftMax) {
leftMax = height[left];
} else {
result += (leftMax - height[left]);
}
++left;
} else {
if (height[right] > rightMax) {
rightMax = height[right];
} else {
result += (rightMax - height[right]);
}
--right;
}
}
return result;
}
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/**
* @author D瓜哥 · https://www.diguage.com
* @since 2024-07-31 17:27:46
*/
public int trap(int[] height) {
int result = 0;
Deque<Integer> stack = new LinkedList<>();
stack.push(0);
for (int r = 1; r < height.length; r++) {
// 单调递减，则将小的元素都出栈
while (!stack.isEmpty() && height[stack.peek()] < height[r]) {
// 递减栈，栈顶最小。与下一个元素比，栈顶小；
// 上面判断条件，栈顶与当前位置元素比，也是栈顶小
int mid = stack.pop();
if (!stack.isEmpty()) {
int l = stack.peek();
// 高h 取两边最小的一个。
// l 是现栈顶元素大，mid 是前栈顶元素最小，当前元素比 mid 大，
// 所以，形成了一个凹槽，可以接水
int h = Math.min(height[l], height[r]) - height[mid];
int w = r - l - 1;
int area = h * w;
result += area;
}
}
stack.push(r);
}
return result;
}
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/**
* @author D瓜哥 · https://www.diguage.com
* @since 2024-08-15 19:07:40
*/
public int trap(int[] height) {
int result = 0;
int left = 0, right = height.length - 1;
int lMax = 0, rMax = 0;
while (left < right) {
lMax = Math.max(lMax, height[left]);
rMax = Math.max(rMax, height[right]);
if (lMax < rMax) {
result += lMax - height[left];
left++;
} else {
result += rMax - height[right];
right--;
}
}
return result;
}