【LeetCode】No.1672. Richest Customer Wealth

LeetCode】No.1672. Richest Customer Wealth"/>

【LeetCode】No.1672. Richest Customer Wealth

题目链接： /

1. 题目介绍（最有钱的人）

You are given an m x n integer grid accounts where accounts[i][j] is the amount of money the i​​​​​​​​​​​​​​​ customer has in the j​​​​​​​​​​​​​​​ bank. Return the wealth that the richest customer has.

【Translate】： 你将获得一个m x n个整数网格accounts，其中accounts[i][j]是第i个客户在第j家银行的钱的数额。请你返回最富有的customer拥有的财富是多少。

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.

【Translate】： 一个customer的财富是指他所有银行账户上的钱的数量。最富有的customer是指拥有最多钱的客户。

测试用例：



约束：

2. 题解

2.1 普普通通小冒泡

  这一题与之前的那道【No.1337. The K Weakest Rows in a Matrix】很相似，No.1337.要求的是在一个m × n的二进制矩阵中找出k个最weak行的索引。相比它，这一题更加简单，它只需要返回最有钱的customer即可，就少了全排序的过程，只需要找出最大的那一个即可。所以，普通的解法就非常简单了，首先遍历数组，将每一行数字的总和存入一个数组；然后对这个数组进行一轮冒泡排序选出最大的那个即可。

    public int maximumWealth1(int[][] accounts) {int m = accounts.length;int n = accounts[0].length;int[] wealth = new int[m];// 遍历数组，并将每行的总和存入wealth[]for(int i = 0; i < m; i++){for(int j = 0; j < n; j++){wealth[i] += accounts[i][j]; }}// 一轮冒泡排序选最大for(int i = 0; i < m-1; i++){if(wealth[i] > wealth[i+1]){int tmp = wealth[i];wealth[i] = wealth[i+1];wealth[i+1] = tmp;}}    return wealth[m-1];     }

2.2 官方题解（Solution）

  官方题解就显得更加的简洁、高效；直接采用一个整形变量maxWealthSoFar用来记录最大值，然后整个大循环结束后返回即可。

    public int maximumWealth2(int[][] accounts) {// Initialize the maximum wealth seen so far to 0 (the minimum wealth possible)int maxWealthSoFar = 0;// Iterate over accountsfor (int[] account : accounts) {// For each account, initialize the sum to 0int currCustomerWealth = 0;// Add the money in each bankfor (int money : account) {currCustomerWealth += money;}// Update the maximum wealth seen so far if the current wealth is greater// If it is less than the current summaxWealthSoFar = Math.max(maxWealthSoFar, currCustomerWealth);}// Return the maximum wealthreturn maxWealthSoFar;}

3. 可参考

[1] 冒泡排序