Container With Most Water 求容器里能放最多的水 @LeetCode

本文介绍了如何通过O(n)双指针法解决经典的容器最多水问题，详细解释了短板效应原理及算法实现过程。

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思路：

O(n) 双指针计算面积，短板效应

package Level3;

/**
 * Container With Most Water
 * 
 *  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.
 *
 */
public class S11 {

	public static void main(String[] args) {

	}
	
	// O(n) 双指针计算面积
	public int maxArea(int[] height) {
		int i = 0, j = height.length-1;
		int max = Integer.MIN_VALUE;
		
		while(i < j){
			// 计算面积的公式
			int area = (j-i) * Math.min(height[i], height[j]);
			// 保存最大面积
			max = Math.max(max, area);
			
			// 因为短板效应，所以只要有一块短板，其他再长也没用，
			// 所以对于短板指针，要寻找下一块板
			if(height[i] < height[j]){
				i++;
			}else{
				j--;
			}
		}
		return max;
    }

}

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