【leetcode】11. Container With Most Water最大水容器(中等難度)
阿新 • • 發佈:2019-02-19
題目:
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 and n is at least 2.
Example:
Input: [1,8,6,2,5,4,8,3,7]
Output: 49
解法一:
對所有的數進行不重複的的組隊遍歷,並寫出一個函式,把每對數放進去計算出大小儲存,直至所有的數遍歷完成,則找到了最大值。
//窮舉法,效能比較差。 public static int maxArea(int[] height) { int sum = 0; if (height.length == 1 || height.length ==0) return 0; for (int i=0;i<height.length;i++){ //利用窮舉法,計算海水的面積,計算出最大的值並儲存下來。 for (int j=i+1;j<height.length;j++){ sum = Math.max(sum,calateSum(height,i,j)); } } return sum; } //計算兩個數字之間的海水大小 public static int calateSum(int[] height,int a,int b){ return Math.min(height[a],height[b])*Math.abs(a-b); }
解法二:
為了減少時間複雜度,則想到了設定兩個指標,類似於快速排序,頭指標和尾指標,計算兩個指標之間的海水容納量,並指標往對應陣列值比較小的地方移動,兩個指標逐漸匯合,直至相鄰,計算結束。
public static int maxArea1(int[] height){ int maxArea=0;int l = 0; int r = height.length - 1;//定義左邊界和右邊界 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; }