学步园

知道是求连续最大子数组后就简单了。但是注意边界条件，如果最大子数组之和＜0，那就不要交易了， 返回0.

public class Solution {
    public int maxProfit(int[] prices) {
    	
    	if(prices.length < 2)
    		return 0;
        
    	int n = prices.length;
    	int[] diffs = new int[n];
        
        for(int i=0;i<n-1;i++)
        	diffs[i] = prices[i+1] - prices[i];
        
        int csum = 0;
        int maxsum = 0;
        
        for(int i=0;i<n-1;i++)
        {
        	csum += diffs[i];
        	if(csum > maxsum)
        		maxsum = csum;
        	
        	if(csum < 0)
        		csum = 0;
        }
        
        return maxsum;
    }
}