LeetCode — 11. Container With Most Water

題目連結：https://leetcode.com/problems/container-with-most-water/description/

先看題目

You are given an integer array height of length n. There are n vertical lines drawn such that the two endpoints of the ith line are (i, 0) and (i, height[i]).

Find two lines that together with the x-axis form a container, such that the container contains the most water.

Return the maximum amount of water a container can store.

Notice that you may not slant the container.

Example 1:

Input: height = [1,8,6,2,5,4,8,3,7]
Output: 49
Explanation: The above vertical lines are represented by array [1,8,6,2,5,4,8,3,7]. In this case, the max area of water (blue section) the container can contain is 49.

Example 2:

Input: height = [1,1]
Output: 1

Constraints:

• n == height.length
• 2 <= n <= 105
• 0 <= height[i] <= 104

解題思路

管他的，先暴力解看看，沒有超時就不管了XD

• 找到最高的線條的高度：maxHeight
• 從高度 0 到 maxHeight 依序找到最外圍的線條並計算容量
• 取最高的容量輸出

這題真的很簡單（可能 Hard 寫太多，Medium 的題目瞬間變的很簡單）：

class Solution {
public:
    int maxArea(vector<int>& height) {
        int n = height.size();
        int maxHeight = 0;
        int maxWater = 0;
        for (int i = 0; i < n; i++) {
            maxHeight = max(maxHeight, height[i]);
        }
        for (int i = 0; i <= maxHeight; i++) {
            int leftIndex = -1;
            int rightIndex = -1;
            for (int j = 0; j < n; j++) {
                if (height[j] >= i) {
                    leftIndex = j;
                    break;
                }
            }
            for (int j = n - 1; j >= 0; j--) {
                if (height[j] >= i) {
                    rightIndex = j;
                    break;
                }
            }
            if (leftIndex == rightIndex || leftIndex == -1) {
                break;
            }
            maxWater = max(maxWater, i * (rightIndex - leftIndex));
        }
        return maxWater;
    }
};

執行結果

最後壓線 AC 通過，但耗時超高。後來看其他人的解法，他們是以遍歷線條為出發點，由外向內取較高的線條計算最大容量。寫競程的題目真的可以訓練邏輯思考能力。

這題的解題紀錄到這邊就結束了喔！如果有任何疑問或建議，歡迎來信詢問：sunyipingtw@icloud.com。記得追蹤加按讚～