best time to Buy and Sell Stock Leetcode

Say you has an array for which the i-th element is the price of a given-stock on day I.

If you were-permitted-to-complete at most one transaction (ie, buy one and sell one share of the stock), design an AL Gorithm to find the maximum profit.

Example 1:

Input: [7, 1, 5, 3, 6, 4]output:5max. difference = 6-1 = 5 (Not 7-1 = 6, as selling-price needs-be-larger than buying price)

Example 2:

Input: [7, 6, 4, 3, 1]output:0in this case, no transaction was done, i.e. Max Profit = 0.

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This problem is not seen in the beginning of dynamic planning ... It's good to see it. So pay attention to recognition.

 Public classSolution { Public intMaxprofit (int[] prices) {        if(Prices = =NULL|| Prices.length = = 0) {            return0; }        int[] res =New int[Prices.length]; intLow = Prices[0];res[0] = 0;  for(inti = 1; i < prices.length; i++) {            if(Prices[i] <Low ) { Low=Prices[i]; } Res[i]= Math.max (Res[i-1], prices[i]-Low ); }        returnRes[prices.length-1]; }}

Of course, this code can also be optimized to optimize the space of O (1)

 Public classSolution { Public intMaxprofit (int[] prices) {        if(Prices = =NULL|| Prices.length = = 0) {            return0; }        intMax = 0; intLow = Prices[0];  for(inti = 1; i < prices.length; i++) {            if(Prices[i] >Low ) {Max= Math.max (max, Prices[i]-Low ); } Else{ Low=Prices[i]; }                    }        returnMax; }}

Notice the conditional judgment of the second method.

best time to Buy and Sell Stock Leetcode