LeetCode121买卖股票的最佳时机

 方法一：直接遍历，时间复杂度为O(n^2)

int maxProfit(int* prices, int pricesSize) {
    //直接遍历    
    int max = 0;
    int i,j;
    for(i = 0; i < pricesSize - 1; i++)
        for(j = i+1; j < pricesSize; j++)
        {
            if(prices[j] - prices[i] > max)
                max = prices[j] - prices[i];
        }
    return max;
}

方法二：一次遍历，min和maxprices分别表示当前位置之前的最小值及此时最大利润

int maxProfit(int* prices, int pricesSize) {
    int maxprofit = 0;
    int i;
    int min = prices[0];
    for(i = 0; i < pricesSize; i++)
    {
        if(min > prices[i])
            min = prices[i];
        else if(prices[i] - min > maxprofit)
            maxprofit = prices[i] - min;
    }
    return maxprofit;
}