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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.
Analyse: Set two variables, left and right to represent the leftmost and rightmost element respectively. We increase left and decrease right gradually. If the moving items are smaller than height[left] and height[right], then we don‘t need to change the value of left and right. Otherwise, we change the values and compare the area with the former computed one.
Runtime: 32ms.
1 class Solution { 2 public: 3 int maxArea(vector<int>& height) { 4 if(height.empty()) return 0; 5 6 int left = 0, right = height.size() - 1; 7 int result = 0; 8 while(left <= right){ 9 int h = min(height[left], height[right]); 10 result = max(result, h * (right - left)); 11 12 if(height[left] > height[right]){ 13 int index = right; 14 while(left <= index && height[index] <= height[right]) 15 index--; 16 right = index; 17 } 18 else{ 19 int index = left; 20 while(index <= right && height[index] <= height[left]) 21 index++; 22 left = index; 23 } 24 } 25 return result; 26 } 27 };
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原文地址：http://www.cnblogs.com/amazingzoe/p/4777511.html
