leetcode 11. Container With Most Water-优快云博客

本文介绍了一种求解最大面积水容器问题的高效算法。该算法通过预处理和双指针技巧，在O(n)的时间复杂度内找到由给定高度数组所能构成的最大矩形面积。文章提供了详细的代码实现及解析。

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class Solution {
public:
    int ans(vector<int>&height){
        if(height.size() < 2) return 0;
        vector<int> up;
        up.push_back(0);
        int ans = 0;
        for(int i = 1;i < height.size(); i++){
            if(height[i] > height[up[up.size()-1]])
                up.push_back(i);
            int low = 0,high = up.size()-1;
            while(low <= high){
                int mid = (low+high)/2;
                if(height[up[mid]] < height[i]) low = mid+1;
                else high = mid-1;
            }
            if(low != up.size()){
                ans = max(ans,(i-up[low])*height[i]);
            }
        }
        return ans;
    }
    int maxArea(vector<int>& height) {
        int ans1 = ans(height);
        int n= height.size();
        for(int i = 0;i < n/2; i++){
            swap(height[i],height[n-i-1]);
        }
        ans1 = max(ans1,ans(height));
        return ans1;
    }
};

class Solution {
public:
    int up[100000];
    int down[100000];
    int ans(vector<int>&height){
        if(height.size() < 2) return 0;
        int top = 0,tail=0;
        up[top++] = 0;
        int ans = 0;
        for(int i = 1;i < height.size(); i++){
            if(height[i] > height[up[top-1]])
                up[top++] = i;
        }
        int f = -1;
        int top2 = 0,tail2=0;
        for(int i = height.size()-1;i>0;i--){
            if(height[i] <= f) continue;
            f = height[i];
            down[top2++] = i;
            while(up[top-1] >= i && top > tail) top--;
            while(height[up[tail]] < f && tail < top) tail++;
            if(tail < top && ans < f * (i-up[tail]))
                ans = f*(i-up[tail]);
        }
        f = -1;
        for(int i = 0;i < height.size()-1; i++){
            if(height[i] <= f) continue;
            f = height[i];
            while(down[top2-1]<=i&&top2>tail2)top2--;
            while(height[down[tail2]] < f && tail2 < top2) tail2++;
            if(tail2 < top2){
                ans = max(ans,f*(down[tail2]-i));
            }
        }
        return ans;
    }
    int maxArea(vector<int>& height) {
        int ans1 = ans(height);
        return ans1;
    }
};