37 lines
1.3 KiB
JavaScript
37 lines
1.3 KiB
JavaScript
/**
|
||
* https://leetcode.cn/problems/best-time-to-buy-and-sell-stock/description/?envType=study-plan-v2&envId=top-interview-150
|
||
* @param {number[]} prices
|
||
* @return {number}
|
||
* 暴力解法：遍历整个数组，找到买入点和所有卖出点的利润，如果利润大于当前利润就更新最大利润，最后返回最大利润,会超时
|
||
*/
|
||
|
||
function f1(prices) {
|
||
let maxProfit = 0;
|
||
const size = prices.length;
|
||
for (let i = 0; i < size; i++) {
|
||
for (let j = i + 1; j < size; j++) {
|
||
if (prices[j] - prices[i] > maxProfit) maxProfit = prices[j] - prices[i];
|
||
}
|
||
}
|
||
|
||
return maxProfit;
|
||
}
|
||
|
||
/**
|
||
* @param {number[]} prices
|
||
* @return {number}
|
||
*单次遍历法，定义两个变量，一个maxProfit = 0；一个minPrice=Infinity，maxProfit和上面的暴力法是一样的，保存最大利润，maxProfit
|
||
*表示的是买入的最小价格，买入的越小后面如果遇到大于它的价格利润就越大，如果利润比maxProfit还大就可以更新maxProfit了，最后
|
||
*返回即可
|
||
*/
|
||
function f2(prices) {
|
||
let maxProfit = 0;
|
||
let minPrice = Infinity;
|
||
for (let i = 0; i < prices.length; i++) {
|
||
if (prices[i] < minPrice) {
|
||
minPrice = prices[i];
|
||
} else if (prices[i] - minPrice > maxProfit) { maxProfit = prices[i] - minPrice; }
|
||
}
|
||
return maxProfit;
|
||
}
|