題目鏈接：http://oj.leetcode.com/problems/container-with-most-water/

第一個解法是用兩個數組雙向記錄。方法的時間復雜度是O(n+max)。這里的max是height的最大值。當時沒有想到用雙向遍歷的方法去做。【貌似第一遍leetcode中還從來沒有用過雙向遍歷】

1 class Solution { 2 public: 3 int maxArea(vector<int> &height) { 4 // Start typing your C/C++ solution below 5 // DO NOT write int main() function 6 int max=-1; 7 int * ar; 8 int n = height.size(); 9 ar = new int[n]; 10 vector<int>::iterator ii; 11 int i=0; 12 for (ii= height.begin(); ii!=height.end();ii++, i++){ 13 if (*ii>max) 14 max = *ii; 15 ar[i] = *ii; 16 } 17 int * f, *g; 18 f = new int[max+1]; 19 g = new int [max+1]; 20 for (i=0; i<=max; i++){ 21 f[i]=-1; 22 g[i]=-1; 23 } 24 int j; 25 for (i=0; i<n; i++){ 26 j = ar[i]; 27 while(j>=0 && f[j]==-1){ 28 f[j] = i; 29 j--; 30 } 31 } 32 for (i=n-1; i>=0; i--){ 33 j = ar[i]; 34 while (j>=0 && g[j]==-1){ 35 g[j] = i; 36 j--; 37 } 38 } 39 int area = 0; 40 for (i=0; i<n; i++){ 41 if (g[ar[i]]>i){ 42 int ta = (g[ar[i]]-i)*ar[i]; 43 if (ta>area) 44 area = ta; 45 } 46 } 47 for (i=n-1; i>=0; i--){ 48 if (f[ar[i]]<i){ 49 int ta = (i-f[ar[i]])*ar[i]; 50 if (ta>area) 51 area = ta; 52 } 53 } 54 delete []ar; 55 delete []f; 56 delete []g; 57 return area; 58 } 59 };

?雙向遍歷的算法倒是很簡單：

1 class Solution { 2 public: 3 int maxArea(vector<int> &height) { 4 // Start typing your C/C++ solution below 5 // DO NOT write int main() function 6 int l=0, r = height.size()-1; 7 int area = 0; 8 while (l<r){ 9 int t = (r-l)*min(height[l], height[r]); 10 if (t>area) 11 area = t; 12 if (height[l]<height[r]) 13 l++; 14 else r--; 15 } 16 return area; 17 } 18 };

不僅復雜度降為O(n)，代碼量更是大大的少了。

至于證明。你懂的。

?

轉載于:https://www.cnblogs.com/zhanghs/archive/2013/03/26/2983624.html

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