6. 42_接雨水
![img](https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2018/10/22/rainwatertrap.png)
/*
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)
*/
/*
暴力解, O(n²),O(1)
*/
class Solution {
int trap(int[] height)
{
int ans = 0;
int size = height.length;
for (int i = 1; i < size - 1; i++) {
int leftMax = 0, rightMax = 0;
for (int j = i; j >= 0; j--) {
leftMax = Math.max(leftMax, height[j]);
}
for (int j = i; j < size; j++) {
rightMax = Math.max(rightMax, height[j]);
}
ans += Math.min(leftMax, rightMax) - height[i];
}
return ans;
}
}
/*
动态编程 O(n),O(n)
*/
class Solution {
public int trap(int[] height)
{
if (height == null || height.length == 0)
return 0;
int ans = 0;
int len = height.length;
int[] leftMax = new int[len];
int[] rightMax = new int[len];
leftMax[0] = height[0];
for (int i = 1; i < len; i++) {
leftMax[i] = Math.max(height[i], leftMax[i - 1]);
}
rightMax[len - 1] = height[len - 1];
for (int i = len - 2; i >= 0; i--) {
rightMax[i] = Math.max(height[i], rightMax[i + 1]);
}
for (int i = 1; i < len - 1; i++) {
ans += Math.min(leftMax[i], rightMax[i]) - height[i];
}
return ans;
}
}
/*
栈的应用, O(n),O(n)(单调递减栈)
*/
class Solution {
int trap(int[] height)
{
int ans = 0, cur = 0;
Stack<Integer> st = new Stack<>();
while (cur < height.length) {
while (!st.isEmpty() && height[cur] > height[st.peek()]) {
int top = st.peek();
st.pop();
if (st.isEmpty())
break;
int distance = cur - st.peek() - 1;
int boundedHeight = Math.min(height[cur], height[st.peek()]) - height[top];
ans += distance * boundedHeight;
}
st.push(cur++);
}
return ans;
}
}
/*
双指针,O(n),O(1)
*/
class Solution {
int trap(int[] height)
{
int left = 0, right = height.length - 1;
int ans = 0;
int leftMax = 0, rightMax = 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;
}
}