Leetcode 42. Trapping Rain Water

描述

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.

For example,

Given [0,1,0,2,1,0,1,3,2,1,2,1], return 6.

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!

思路

第一种是找水位，具体做法是对于每个点，往左往右分别找出最高值，然后计算取较小者，这就是水位高度。减去当前点的高度就是水的体积。

class Solution {
public int trap(int[] height){
int ans = 0;
for (int i = 1; i < height.length; i++){
int left = 0, right = 0;
for(int j = i; j >= 0; j--) {
left = Math.max(left, height[j]);
}
for(int j = i; j < height.length; j++){
right = Math.max(right, height[j]);
}
ans += Math.min(left, right) - height[i];
}
return ans;
}
}

我们发现上面的算法，有着很多重复计算，只要把结果缓存一下，就可以变成O（N）

public int trap(int[] height){
if (height == null)
return 0;
int ans = 0;
int size = height.length;
int[] left = new int[size];
int[] right = new int[size];
left[0] = height[0];
for (int i = 1; i < size; i++){
left[i] = Math.max(height[i], left[i-1]);
}
right[size - 1] = height[size - 1];
for (int i = size - 2; i >= 0; i--){
right[i] = Math.max(height[i], right[i+1]);
}
for (int i = 1; i < size - 1; i++){
ans += Math.min(left[i], right[i]) - height[i];
}
return ans;
}

时间复杂度肯定是最优了，之后只能优化空间复杂度了。

public int trap(int[] height){
int left = 0;
int right = height.length - 1;
int leftMax = 0, rightMax = 0;
int ans = 0;
while (left < right){
if (height[left] < height[right]){
if (height[left] >= leftMax)
leftMax = height[left];
else
ans += leftMax - height[left];
left++;
}else{
if (height[right] >= rightMax)
rightMax = height[right];
else
ans += rightMax - height[right];
right--;
}
}
return ans;
}