leetcode: Container With Most Water-优快云博客

本文针对LeetCode上的容器盛水问题进行了解析，利用木桶原理寻找最大容积的策略。通过双指针从两端向中间逼近的方式，逐步优化解决方案。

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

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.

思路：

基于木桶原理，容积是由短板决定。从左右两头往当中走，容积等于(right - left) * min(height[left], height[right])。如果左边高度较低的话，右边再往前移是不可能获得更大的容积，只有左边往后移直到height[left]能取得更大的值。

 1 class Solution {
 2 public:
 3     int maxArea(vector<int> &height) {
 4         int width = height.size(), left = 0, right = width - 1;
 5         int max = 0;
 6         
 7         while (left < right) {
 8             int vol = min(height[left], height[right]) * (right - left);
 9             
10             if (vol > max) {
11                 max = vol;
12             }
13             
14             int next;
15             if (height[left] > height[right]) {
16                 next = right - 1;
17                 while ((next > left) && (height[next] <= height[right])) {
18                     --next;
19                 }
20                 right = next;
21             }
22             else {
23                 next = left + 1;
24                 while ((next < right) && (height[next] <= height[left])) {
25                     ++next;
26                 }
27                 left = next;
28             }
29         }
30         
31         return max;
32     }
33 };