接雨水

11. Container With Most Water
42. Trapping Rain Water

11. Container With Most Water

You are given an integer array height of length n. There are n vertical lines drawn such that the two endpoints of the ith line are (i, 0) and (i, height[i]).

Find two lines that together with the x-axis form a container, such that the container contains the most water.

Return the maximum amount of water a container can store.

Notice that you may not slant the container.

Example 1:

Input: height = [1,8,6,2,5,4,8,3,7]
Output: 49
Explanation: The above vertical lines are represented by array [1,8,6,2,5,4,8,3,7]. In this case, the max area of water (blue section) the container can contain is 49.

Example 2:

Input: height = [1,1]
Output: 1

Constraints:

n == height.length
2 <= n <= 10^5
0 <= height[i] <= 10^4

思路

双指针，左低移左，右低移右

The two-pointer technique starts with the widest container and moves the pointers inward based on the comparison of heights.

Increasing the width of the container can only lead to a larger area if the height of the new boundary is greater. By moving the pointers towards the center, we explore containers with the potential for greater areas.

C++ 解法

class Solution {
public:
    int maxArea(vector<int>& height) {
        int left = 0;
        int right = height.size() - 1;
        int maxArea = 0;

        while (left < right) {
            int currentArea = min(height[left], height[right]) * (right - left);
            maxArea = max(maxArea, currentArea);

            if (height[left] < height[right]) {
                left++;
            } else {
                right--;
            }
        }

        return maxArea;
    }
};

Java 解法

class Solution {
    public int maxArea(int[] height) {
        int result = 0;
        int left = 0; 
        int right = height.length - 1;
        while(left <= right){
            result = Math.max(result, Math.min(height[right],height[left])*(right-left));
            if(height[right] < height[left]) right--;
            else left++;
        }
        return result;
    }
}

Python 解法

class Solution:
    def maxArea(self, height: List[int]) -> int:
        left = 0
        right = len(height) - 1
        maxArea = 0

        while left < right:
            currentArea = min(height[left], height[right]) * (right - left)
            maxArea = max(maxArea, currentArea)

            if height[left] < height[right]:
                left += 1
            else:
                right -= 1

        return maxArea

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 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

思路

一共三种解法：双指针、单调栈和动态规划。

C++ 解法

单调栈

class Solution {
public:
    int trap(vector<int>& height) {
        if(height.size() <= 2) return 0;
        int result = 0;
        stack<int> st;
        st.push(0);
        for(int i = 1; i < height.size(); i++){
            if(height[i] < height[st.top()]){
                st.push(i);
            }else if(height[i] == height[st.top()]){
                st.push(i);
            }else{
               while(!st.empty() && height[i] > height[st.top()]){
                    int mid = st.top();
                    st.pop();
                    if(!st.empty()){
                        int h = min(height[i], height[st.top()]) - height[mid];
                        int w = i - st.top() - 1;
                        result += h * w;
                    }
               }
               st.push(i);
            }
        }
        return result;
    }
};

Java 解法

双指针

class Solution {
    public int trap(int[] height) {
        int i=0,left_max=height[0];
        int sum=0;
        int j=height.length-1,right_max=height[j];
        while(i < j){
            if(left_max <= right_max){
                sum += left_max - height[i];
                i++;
                left_max = Math.max(left_max, height[i]);
            }else{
                sum += right_max - height[j];
                j--;
                right_max = Math.max(right_max, height[j]);
            }
        }
        return sum;
    }
}

Python 解法

双指针

class Solution:
    def trap(self, height: List[int]) -> int:
        i = 0
        left_max = height[0]
        sum = 0
        j = len(height) - 1
        right_max = height[j]
        while i < j:
            if left_max <= right_max:
                sum += left_max - height[i]
                i += 1
                left_max = max(left_max, height[i])
            else:
                sum += right_max - height[j]
                j -= 1
                right_max = max(right_max, height[j])
        return sum

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