[LeetCode] Container With Most Water
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原文地址:http://www.cnblogs.com/remlostime/archive/2012/10/29/2745018.html
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.
类似于2Sum的思想,两边设一个指针,然后计算area,如果height[i] <= height[j],那么i++,因为在这里height[i]是瓶颈,j往里移只会减少面积,不会再增加area。
这是一个贪心的策略,每次取两边围栏最矮的一个推进,希望获取更多的水。
一个不严格的证明:
当height[i] <= height[j]时,为什么是i++,而不是j++来获取可能更多的水?
假设j' > j,之所以j'往左移,是因为存在height[i'] > height[j'] (i’ <= i), 而那时area' = (j' - i') * min(height[i'], height[j']),
因为height[j'] == min(height[i'], height[j']),所以area' = (j' - i') * height[j']。
而i 和 j'构成的面积area = (j' - i) * min(height[i], height[j'])。
area' >= area,所以j不需要往右移。
class Solution {public: int maxArea(vector<int> &height) { // Start typing your C/C++ solution below // DO NOT write int main() function int i = 0; int j = height.size() - 1; int ret = 0; while(i < j) { int area = (j - i) * min(height[i], height[j]); ret = max(ret, area); if (height[i] <= height[j]) i++; else j--; } return ret; }};
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