LeetCode 121 Best Time to Buy and Sell Stock (贪心)
2016-09-04 11:31
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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:
Example 2:
题目链接:https://leetcode.com/problems/best-time-to-buy-and-sell-stock/
题目分析:当前数字比最小值小则更新最小值,否则算差值更新答案
public class Solution {
public int maxProfit(int[] prices) {
int len = prices.length;
if (len == 0) {
return 0;
}
int mi = prices[0], ans = 0;
for (int i = 1; i < len; i ++) {
if(prices[i] < mi) {
mi = prices[i];
}
else {
ans = Math.max(ans, prices[i] - mi);
}
}
return ans;
}
}
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.
题目链接:https://leetcode.com/problems/best-time-to-buy-and-sell-stock/
题目分析:当前数字比最小值小则更新最小值,否则算差值更新答案
public class Solution {
public int maxProfit(int[] prices) {
int len = prices.length;
if (len == 0) {
return 0;
}
int mi = prices[0], ans = 0;
for (int i = 1; i < len; i ++) {
if(prices[i] < mi) {
mi = prices[i];
}
else {
ans = Math.max(ans, prices[i] - mi);
}
}
return ans;
}
}
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