11-Container With Most Water

【题目】

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.

【analyze】

1.穷举的方法时间复杂度为O(n)

2.若从两边逼近并记录最大值，令left=0,right=length-1;若height[left]>height[right]，则right--;反之，则left++。

　为什么能这么做呢？需要用到反证思想，假设i-j取到最大值，则i之前都要小于height[i]

3.还可改进下，不用每次都减一

【算法】

v1:
public class Solution {
    public int maxArea(int[] height) {
        int res=0;
        int l=0;
        int r=height.length-1;
        while(l<r) {
            int w=r-l;
            int h=Math.min(height[l],height[r]);
            res=Math.max(res,w*h);
            if(height[l]<height[r])
                l++;
            else 
                r--;
        }
        return res;
    }
}

v2:
public class Solution {
    public int maxArea(int[] height) {
        int res=0;
        int l=0;
        int r=height.length-1;
        while(l<r) {
            int w=r-l;
            int h=Math.min(height[l],height[r]);
            res=Math.max(res,w*h);
            if(height[l]<height[r]) {
                int temp=l;
                while(temp<r&&height[temp]<=height[l])
                    temp++;
                l=temp;
            }
            else {
                int temp=r;
                while(temp>l&&height[temp]<=height[r])
                    temp--;
                r=temp;
            }
        }
        return res;
    }
}