Container With Most Water--LeetCode

Container With Most Water

(原题链接：点击打开链接)

Given n non-negative integers a₁,a₂, ..., a_n, where each represents a point at coordinate (i,a_i). n vertical lines are drawn such that the two endpoints of linei is at (i, a_i) 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 and n is at least 2.

Solution:

超时（时间复杂度为O(N^2)）：

class Solution {public:    int maxArea(vector<int>& height) {        int maxWater = 0;        int localheight = 0;        for (int a = 0; a < height.size() - 1; a++)        {            for (int b = a + 1; b < height.size(); b++)            {                localheight = height[a] > height[b] ? height[b] : height[a];                int localWater = localheight * (b - a);                maxWater = maxWater > localWater ? maxWater : localWater;            }        }        return maxWater;    }};

我们从上面方法中可以得到a总是小于b的。而且中间会做很多无用功。所以我们可以尝试一下将方法改良，变成a，b分别指示两端，并且向中间移动，而不是双层循环。此时时间复杂度变为为O(N):

当height[a] <= localheight && a < b时，此时假设是a变大引起的改变，b-a减少，height[a]也不能使localheight增大，即localheight不变或减少，此时localWater肯定是减少的，没有意义，所以就不用再计算这个值，它是不会使maxWater发生变化的。

当height[b] <= localheight && a < b时，此时假设是b变小引起的改变，b-a减少，height[b]也不能使localheight变大，所以此时localWater肯定是减少的，没有意义，同样不用计算。

a，b向中间移动直至a == b

class Solution {public:    int maxArea(vector<int>& height) {        int maxWater = 0;        int localheight = 0;        int a = 0;        int b = height.size() - 1;        while (a < b)        {            localheight = height[a] > height[b] ? height[b] : height[a];            int localWater = localheight * (b - a);            maxWater = maxWater > localWater ? maxWater : localWater;            while (height[a] <= localheight && a < b) a++;            while (height[b] <= localheight && a < b) b--;        }        return maxWater;    }};