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.
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題目

給定 n 個(gè)非負(fù)整數(shù) a1, a2, ..., an, 每個(gè)數(shù)代表了坐標(biāo)中的一個(gè)點(diǎn) (i, ai)。畫 n 條垂直線，使得 i 垂直線的兩個(gè)端點(diǎn)分別為(i, ai)和(i, 0)。找到兩條線，使得其與 x 軸共同構(gòu)成一個(gè)容器，以容納最多水。
注意事項(xiàng)
容器不可傾斜。
樣例
給出[1,3,2], 最大的儲(chǔ)水面積是2

分析

用兩根指針一頭一尾，分別計(jì)算面積，然后選取高度小的那邊向中間移動(dòng)，因?yàn)檫@樣才可能存在更大的面積，更新最大的面積，最后就是結(jié)果。

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代碼

public class Solution {
    /**
     * @param heights: an array of integers
     * @return: an integer
     */
    int computeArea(int left, int right,  int[] heights) {
        return (right-left)*Math.min(heights[left], heights[right]);
    }
    
    public int maxArea(int[] heights) {
        //兩根指針，一頭一尾計(jì)算，短的那根指針向中間移動(dòng)，因?yàn)橹挥羞@種情況才存在更大的情況
        int left = 0, right = heights.length-1;
        int res = 0;
        
        while(left < right) {
            res = Math.max(res, computeArea(left,right,heights));
            if(heights[left] < heights[right]) {
                left++;
                //如果小于則不必要計(jì)算，直接跳過(guò)
                while(heights[left] <= heights[left-1] && left<right)
                    left++;
            }
            else {
                right--;
                while(heights[right] <= heights[right+1] && left <right)
                    right--;
            }
        }
        return res; 
    }
}