LeetCode 2679. Sum in a Matrix
You are given a?0-indexed?2D integer array?nums. Initially, your score is?0. Perform the following operations until the matrix becomes empty:
From each row in the matrix, select the largest number and remove it. In the case of a tie, it does not matter which number is chosen.
Identify the highest number amongst all those removed in step 1. Add that number to your?score.
Return?the final?score.
?
Example 1:
Input: nums = [[7,2,1],[6,4,2],[6,5,3],[3,2,1]]
Output: 15
Explanation: In the first operation, we remove 7, 6, 6, and 3. We then add 7 to our score. Next, we remove 2, 4, 5, and 2. We add 5 to our score. Lastly, we remove 1, 2, 3, and 1. We add 3 to our score. Thus, our final score is 7 + 5 + 3 = 15.
Example 2:
Input: nums = [[1]]Output: 1
Explanation: We remove 1 and add it to the answer. We return 1.
?
給你一個(gè)下標(biāo)從 0 開始的二維整數(shù)數(shù)組 nums 。一開始你的分?jǐn)?shù)為 0 。你需要執(zhí)行以下操作直到矩陣變?yōu)榭眨?/p>
矩陣中每一行選取最大的一個(gè)數(shù)，并刪除它。如果一行中有多個(gè)最大的數(shù)，選擇任意一個(gè)并刪除。
在步驟 1 刪除的所有數(shù)字中找到最大的一個(gè)數(shù)字，將它添加到你的 分?jǐn)?shù) 中。
請(qǐng)你返回最后的 分?jǐn)?shù) 。
可以先將每一行數(shù)組進(jìn)行排序，然后依次求出每一列的最大值，sum求和記錄即可；
Runtime:?16 ms, faster than?88.62%?of?Java?online submissions for?Sum in a Matrix.
Memory Usage:?57 MB, less than?99.68%?of?Java?online submissions for?Sum in a Matrix.
