Leetcode 11. Container with Most Water

算法 Java Leetcode Golang

Find a container which contains the most water.

Question

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.
See it on Leetcode

For example,
int[] a = { 1, 8, 6, 2, 5, 4, 8, 3, 7 },
it should return 49.

Two Pointer Approach

Note

1. You may not slant the container.
2. Brute Force approach may cause TLE.

Solution in Brute Force - TLE

• java

public class Solution {
    public int maxArea(int[] height) {
        int n = height.length, maxArea = 0;
        for (int i = 0; i < n - 1; i++) {
            for (int j = i + 1; j < n; j++) {
                maxArea = Math.max((j - i) * Math.min(height[i], height[j]), maxArea);
            }
        }
        return maxArea;
    }
}

Solution in Java, C++ and Golang

• java
• cpp
• go

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

class Solution {
public:
    int maxArea(vector<int>& height) {
        int maxArea = 0;
        int l = 0, r = height.size() - 1;
        while (l < r) {
            maxArea = fmax(maxArea, fmin(height[l], height[r]) * (r - l));
            if (height[l] < height[r]) {
                l++;
            } else {
                r--;
            }
        }
        return maxArea; 
    }
};

func maxArea(height []int) int {
    var maxArea, l, r int = 0, 0, len(height) - 1
    for l < r {
        switch {
        case height[l] < height[r]:
            if height[l] * (r - l) > maxArea {
                maxArea = height[l] * (r - l)
            }
        default:
            if height[r] * (r - l) > maxArea {
                maxArea = height[r] * (r - l)
            }
        }
        if height[l] < height[r] {
            l++
        } else {
            r--
        }
    }
    return maxArea
}