前面两道Hard类型的题目 等我彻彻底底弄明白了再仔仔细细写一篇先。
这道Medium的题我觉得超级棒的说。出题的人太有水平了，比直方图面积的单调栈要简单，暴力解法会超时，但是O(N)的解法是又精巧又直观，我自己就没想出来，看了提示才知道的。
na1a2an, where each represents a point at coordinate (iainiiai) and (i, 0). Find two lines, which together with x-axis forms a container, such that the container contains the most water.
n
提示：
Initially we consider the area constituting the exterior most lines.
Now, to maximize the area, we need to consider the area between the lines of larger lengths.
since it is limited by the shorter line.
This is done since a relatively longer line obtained by moving the shorter line's pointer might overcome the reduction in area caused by the width reduction.
class Solution { public: int maxArea(vector<int>& height) { typedef vector<int>::iterator it; it begin, end; int maxArea = 0; int length = (int)height.size() - 1; //别忘了减一 begin = height.begin(); end =height.end(); end--; while(length > 0){ maxArea = max(maxArea , min(*begin, *end) * length); length --; if(*begin < *end) begin++; else end--; } return maxArea; } };