|112|[Path Sum][112]|Tree, Depth-first Search|

|118|[Pascal's Triangle][118]|Array|

|119|[Pascal's Triangle II][119]|Array|

|121|[Best Time to Buy and Sell Stock][121]|Array, Dynamic Programmin|

[112]:https://github.com/Blankj/awesome-java-leetcode/blob/master/note/112/README.md

[118]:https://github.com/Blankj/awesome-java-leetcode/blob/master/note/118/README.md

[119]:https://github.com/Blankj/awesome-java-leetcode/blob/master/note/119/README.md

[121]:https://github.com/Blankj/awesome-java-leetcode/blob/master/note/121/README.md

[008]:https://github.com/Blankj/awesome-java-leetcode/blob/master/note/008/README.md

[019]:https://github.com/Blankj/awesome-java-leetcode/blob/master/note/019/README.md

Given an index*k*, return the*k*th row of the Pascal's triangle.

Given an index*k*, return the*k*^th row of the Pascal's triangle.

For example, given*k* = 3,

Return`[1,3,3,1]`.

Say you have an array for which the*i*^th 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:**

**Tags:** Array, Dynamic Programmin

题意是给出一个数组代表每天的股票金额，让你在最多买卖一次的情况下算出最大的收益额，最简单的就是模拟即可，每次记录当前值减去最小值的差值，与上一次的进行比较然后更新最大值即可。

classSolution {

publicintmaxProfit(int[]prices) {

int max=0, minPrice=Integer.MAX_VALUE;

for (int i=0; i< prices.length;++i) {

if (prices[i]< minPrice) minPrice= prices[i];

int delta= prices[i]- minPrice;

if (delta> max) max= delta;

}

return max;

}

}

如果你同我一样热爱数据结构、算法、LeetCode，可以关注我GitHub上的LeetCode题解：[awesome-java-leetcode][ajl]

[title]:https://leetcode.com/problems/best-time-to-buy-and-sell-stock

[ajl]:https://github.com/Blankj/awesome-java-leetcode