LeetCode 题解（72）: Container With Most Water

题目：

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.

题解：

第一种直观的解法，将高度和原始位置记录在一个struct Node里，对每一个Node按照其height升序排序。然后从排序后的最后一个元素，即height最大的元素开始，计算mostWater，当计算当前元素时，只需要看最左边和最右边的比它高的元素，取其中远的那一个元素。用start和end来记录已经计算过的元素中的当前最左位置和当前最右位置，每次计算前先更新start和end。因为需要排序，所以时间复杂度为O(nlgn)。

struct Node {    int index;    int height;    Node(int i, int h):index(i),height(h){}    Node(){}    bool operator < (const Node &t) const {        return height < t.height;    }};class Solution {public:    int maxArea(vector<int> &height) {        if(height.size() <= 1)            return 0;        int mostWater = 0;        vector<Node> node(height.size());        for(int i = 0; i < height.size(); i++) {            node[i].index = i;            node[i].height = height[i];        }        sort(node.begin(), node.end());                int start = node[height.size()-1].index, end = start;        for(int j = height.size() -  2; j >= 0; j--) {            start = min(start, node[j].index);            end = max(end, node[j].index);            mostWater = max(mostWater, max(node[j].height * (node[j].index - start), node[j].height * (end - node[j].index)));        }        return mostWater;    }};

第二种思路：两个指针，头和尾，计算当前mostWater，若头高，减尾指针，若尾高，加头指针。原因是只有这样才有可能增加储水量。时间复杂度O(n)。

C++;

class Solution {public:    int maxArea(vector<int> &height) {        int length = height.size();        if(length <= 1)            return 0;        int start = 0, end = height.size()-1, mostWater = 0;        if(height[start] <= height[end]) {            mostWater = height[start] * (end - start);            start++;        }        else {            mostWater = height[end] * (end - start);            end--;        }        while(start < end) {            if(height[start] <= height[end]) {                mostWater = max(mostWater, height[start] * (end - start));                start++;            }            else {                mostWater = max(mostWater, height[end] * (end - start));                end--;            }        }        return mostWater;    }};

java：

public class Solution {    public int maxArea(int[] height) {        if(height.length <= 1)            return 0;        int mostWater = 0, start = 0, end = height.length - 1;        while(start < end) {            mostWater = Math.max(mostWater, Math.min(height[start], height[end]) * (end-start));            if(height[start] <= height[end])                start++;            else                end--;        }        return mostWater;    }}

Python:

class Solution:    # @return an integer    def maxArea(self, height):        if len(height) <= 1:            return 0        mostWater = 0        start = 0        end = len(height) - 1        while start < end:            mostWater = max(mostWater, min(height[start], height[end]) * (end - start))            if height[start] <= height[end]:                start += 1            else:                end -= 1        return mostWater