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## Problem

Given n non-negative integers representing an elevation map where the width of each bar is 1, compute how much water it can trap after raining.

Example 1:

Input: height = [0,1,0,2,1,0,1,3,2,1,2,1]
Output: 6
Explanation: The above elevation map (black section) 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.


Example 2:

Input: height = [4,2,0,3,2,5]
Output: 9


Constraints:

• n == height.length
• $1 <= n <= 2 * 10^4$
• $0 <= height[i] <= 10^5$

## Code

r[i] = min(max(h[0~i]), max(h[i~n-1])) - h[i]

class Solution {
public int trap(int[] height) {
int len = height.length;

int[] leftMax = new int[len];
int[] rightMax = new int[len];

for(int i = 0; i < len; i++) {
if(i == 0) {
leftMax[i] = height[i];
} else {
leftMax[i] = Math.max(height[i], leftMax[i - 1]);
}
}

for(int i = len - 1; i >= 0; i--) {
if(i == len - 1) {
rightMax[i] = height[i];
} else {
rightMax[i] = Math.max(height[i], rightMax[i + 1]);
}
}

int res = 0;
for(int i = 0; i < len; i++) {
int max = Math.min(leftMax[i], rightMax[i]);
res += max - height[i];
}

return res;
}
}


leftMax 和 rightMax 都是单调的

class Solution {
public int trap(int[] height) {
int left = 0;
int right = height.length - 1;
int res = 0;

int leftMax = 0;
int rightMax = 0;

while(left < right){
if(height[left] < height[right]){
leftMax = Math.max(leftMax, height[left]);
res += leftMax - height[left];
left++;
} else {
rightMax = Math.max(rightMax, height[right]);
res += rightMax - height[right];
right--;
}
}
return res;
}
}