LeetCode 11. Container With Most Water C++

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Given n non-negative integers a1, a2, ..., an , where each represents a point at coordinate (i, ai). n vertical lines are drawn such that the two endpoints of line i is at (i, ai) and (i, 0). Find two lines, which together with x-axis forms a container, such that the container contains the most water.

Note: You may not slant the container and n is at least 2.

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:

Input: [1,8,6,2,5,4,8,3,7]
Output: 49

方法一：暴力求解

class Solution {
public:
    int maxArea(vector<int>& height) {
        int area = 0;
        int sz = height.size();
        for (int i = 0; i < sz; i++) {
            for (int j = sz - 1; j > i; j--) {                
                int tmparea = min(height[i], height[j]) * (j - i);                
                if (area < tmparea)
                    area = tmparea;
            }
        }
        return area;
    }
};

方法二：适当跳过

由方法一知，当两边的一边确定时，即最小那边确定时，遍历的时候（i和j靠近）面积并不会再增大，只会减少。所以可以跳过最小的那边的遍历，使得复杂度降低

class Solution {
public:
    int maxArea(vector<int>& height) {
        int area = 0;
        int sz = height.size();
        int i = 0, j = sz - 1;
        while (i < j) {
            int tmph = min(height[i], height[j]);
            int tmparea = tmph * (j - i);            
            if (area < tmparea)
                area = tmparea;
            if (height[i] < height[j])
                i++;
            else
                j--;            
        }
        return area;
    }
};

方法三：

由二原理更进一步得，跳跃更多遍历，速度打败98%的对手

class Solution {
public:
    int maxArea(vector<int>& height) {
        int area = 0;
        int sz = height.size();
        int i = 0, j = sz - 1;
        while (i < j) {
            int tmph = min(height[i], height[j]);
            int tmparea = tmph * (j - i);            
            if (area < tmparea)
                area = tmparea;
            while (height[i] <= tmph && i < j) i++;
            while (height[j] <= tmph && i < j) j--;
        }
        return area;
    }
};