Description:
Say you have an array for which the ith element is the price of a given stock on day i.
If you were only permitted to complete at most one transaction (ie, buy one and sell one share of the stock), design an algorithm to find the maximum profit.
Example 1:
Input: [7, 1, 5, 3, 6, 4]
Output: 5
max. difference = 6-1 = 5 (not 7-1 = 6, as selling price needs to be larger than buying price)
Example 2:
Input: [7, 6, 4, 3, 1]
Output: 0
In this case, no transaction is done, i.e. max profit = 0.
My code:
/**
* @param {number[]} prices
* @return {number}
*/
var maxProfit = function(prices) {
let maxProfit = 0, len = prices.length;
for(let i = 0; i < len; i++) {
for(let j = i; j < len; j++) {
if(maxProfit < (prices[j] - prices[i])) {
maxProfit = prices[j] - prices[i];
} else {
continue;
}
}
}
return maxProfit;
};
Note: 题目即找出数组中最大的差值(后-前)并返回该差值
