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72 lines
1.8 KiB
Markdown
72 lines
1.8 KiB
Markdown
# [1672. Richest Customer Wealth](https://leetcode.com/problems/richest-customer-wealth/)
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## 题目
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You are given an `m x n` integer grid `accounts` where `accounts[i][j]` is the amount of money the `ith` customer has in the `jth` bank. Return *the **wealth** that the richest customer has.*
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A customer's **wealth** is the amount of money they have in all their bank accounts. The richest customer is the customer that has the maximum **wealth**.
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**Example 1:**
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```
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Input: accounts = [[1,2,3],[3,2,1]]
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Output: 6
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Explanation:1st customer has wealth = 1 + 2 + 3 = 6
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2nd customer has wealth = 3 + 2 + 1 = 6
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Both customers are considered the richest with a wealth of 6 each, so return 6.
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```
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**Example 2:**
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```
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Input: accounts = [[1,5],[7,3],[3,5]]
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Output: 10
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Explanation:
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1st customer has wealth = 6
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2nd customer has wealth = 10
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3rd customer has wealth = 8
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The 2nd customer is the richest with a wealth of 10.
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```
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**Example 3:**
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```
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Input: accounts = [[2,8,7],[7,1,3],[1,9,5]]
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Output: 17
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```
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**Constraints:**
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- `m == accounts.length`
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- `n == accounts[i].length`
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- `1 <= m, n <= 50`
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- `1 <= accounts[i][j] <= 100`
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## 题目大意
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给你一个 m x n 的整数网格 accounts ，其中 accounts[i][j] 是第 i 位客户在第 j 家银行托管的资产数量。返回最富有客户所拥有的 资产总量 。客户的 资产总量 就是他们在各家银行托管的资产数量之和。最富有客户就是 资产总量 最大的客户。
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## 解题思路
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- 简单题。计算二维数组中每个一位数组的元素总和，然后动态维护这些一位数组和的最大值即可。
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## 代码
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```go
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package leetcode
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func maximumWealth(accounts [][]int) int {
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res := 0
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for _, banks := range accounts {
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sAmount := 0
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for _, amount := range banks {
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sAmount += amount
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}
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if sAmount > res {
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res = sAmount
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}
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}
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return res
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}
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``` |