One pass, bug free: 10分鐘
1.一切從簡單考慮:兩種情況，升，降都覆蓋到。
public class Solution {
public int maxProfit(int[] prices) {
//8:45
int min = Integer.MAX_VALUE;
int max = Integer.MIN_VALUE;
int profit = 0, max_profit = 0;
int length = prices.length;
for (int i = 0; i < length; i++){
if(prices[i] < min){
min = prices[i];
max = prices[i];
profit = max - min;
}
else if(prices[i] > max){
max = prices[i];
profit = max - min;
}
max_profit = Math.max(max_profit, profit);
}
return max_profit;
}
}
解法二:
以上解法并不好，沒有體現(xiàn)DP的思想不利于擴(kuò)展
DP解法: time:O(n); Space:O(n)
public class Solution {
public int maxProfit(int[] prices) {
if(prices.length == 0) return 0;
int length = prices.length;
int[] left = new int[length];
int min = prices[0];
for(int i = 1; i<length; i++){
min = Math.min(min, prices[i]);
left[i] = Math.max(left[i-1], prices[i] - min);
}
return left[length-1];
}
}
解法三:
將解法二Space降為O(1):
public class Solution {
public int maxProfit(int[] prices) {
if(prices.length == 0) return 0;
int length = prices.length;
int left = 0;
int min = prices[0];
for(int i = 1; i<length; i++){
min = Math.min(min, prices[i]);
left = Math.max(left, prices[i] - min);
}
return left;
}
}
