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Leetcode - GasStation

时间:2014-10-12 06:25:47      阅读:210      评论:0      收藏:0      [点我收藏+]

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An kind of interesting problem.  Use an array arr[] to store how many gas are left if we travel from station i to station i+1, arr[i] = gas[i] - cost[i]. Then it becomes an variance of the maximal sub array problem: if the sum of the sub array arr[i....j] < 0, it means we can‘t start at any gas station whose index is within [i..j]. Then we should skip to j+1 and start to scan again. O(n) time and O(n) space:  


public class Solution {
    public int canCompleteCircuit(int[] gas, int[] cost) {
        int[] arr = new int[gas.length];
        int sum = 0;
        for(int i=0;i<gas.length;i++)
        {
        	arr[i] = gas[i] - cost[i];
        	sum+=arr[i];
        }
        
        if(sum < 0) return -1;
        
        int s;
        boolean flag;
        for(int i=0;i<gas.length;i++){
        	if(arr[i] < 0)
        		continue;
        	
        	flag = true;
        	s=0;
        	for(int j=0;j<gas.length;j++)
        	{
        		int ind = (i+j) % gas.length;
        		s += arr[ind];
        		if(s<0)
        		{
        			i = i + j;
        			flag = false;
        			break;
        		}
        	}
        	
        	if(flag == true)
        		return i;
        }
        return -1;
    }
    
    public static void main(String[] args)
    {
    	int[] gas= new int[]{1,2,3,3};
    	int[] cost = new int[]{2,1,5,1};
    	
    	Solution sol = new Solution();
    	System.out.print(sol.canCompleteCircuit(gas, cost));
    	
    }
}


Leetcode - GasStation

标签:style   io   os   ar   java   for   sp   art   on   

原文地址:http://blog.csdn.net/tspatial_thunder/article/details/40013211

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