leet_11

Posted 2020-06-09

题目：https://leetcode.com/problems/container-with-most-water/

Given n non-negative integers a₁, a₂, ..., a_n, where each represents a point at coordinate (i, a_i). n vertical lines are drawn such that the two endpoints of line i is at (i, a_i) 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.

 

n个非负的整数，a1,a2....an, 每一个都代表了一个坐标(i, ai). n条从点(i,ai) 到(i, 0)这样的垂直线。找到这样的两条线，和x轴形成一个容器，可以包含最多的水。

 

 

 

public static int maxArea(int[] height) {

if(null == height){
return 0;
}
int size = height.length;
    int maxS = 0;

    for(int i = 0; i < size; i++){
        for(int j = i + 1;j < size; j++){
            int maxSTemp = Math.min(height[i], height[j]) * (j - i);
            if(maxSTemp > maxS){
                maxS = maxSTemp;
            }
        }
    }

    return maxS;
}

public static int maxArea2(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;
}

 

 

 

Analysis

Initially we can assume the result is 0. Then we scan from both sides. If leftHeight < rightHeight, move right and find a value that is greater than leftHeight. Similarily, if leftHeight > rightHeight, move left and find a value that is greater than rightHeight. Additionally, keep tracking the max value.