【LeetCode Daily Question】[Moderate]134. Gas Station

【LeetCode Daily Question】[Moderate]134. Gas Station

134. Gas Station

134. Gas Station

Algorithm idea: array

topic:

java code-brute force

Violence, enumerate every position, and judge it can travel for a week

class Solution {
    
    
    public int canCompleteCircuit(int[] gas, int[] cost) {
    
    
        //暴力，枚举每一个位置，判断能都行驶一周
		int n = gas.length;
		for (int i = 0; i < n; i++) {
    
    //遍历每一个位置
			int oilMass = 0;
			int  j = i;
			while (oilMass >= 0) {
    
    //判断当前位置能否行驶一周
				oilMass = oilMass + gas[j] - cost[j];
				j = (j + 1) % n;
				if (oilMass >= 0 && j == i) {
    
    //如果可以则返回i
					return i;
				}
			}
		}
		return -1;//不能行驶一周
    }
}

java code-traverse it again

Using, x can reach y and any station before y, but cannot reach the next station of y, then any station in the interval (x,y) cannot reach the next station of y;

explain:

1. Suppose there is x'which is one of the stations [x,y];
2. x can reach each station from x to y, then the sum of [x, x'] and the sum of [x, y] must be >=0,
3. x cannot reach the next stop of y, then the sum of [x, y+1] is superimposed and <0;
4. It can be seen that [x,y+1]-[x, x'] = [x', y+1] <0;
5. Therefore, any station in the [x,y] interval cannot reach the next station of y; (x can be directly jumped to y+1 when traversing, and the middle station does not need to be considered)

class Solution {
    
    
    public int canCompleteCircuit(int[] gas, int[] cost) {
    
    
        //遍历一遍，O(n)
        //利用,x不能到y的下一站,则[x,y]区间内的任意一站无法到达y的下一站
		int n = gas.length;
		int i = 0;
		while (i < n) {
    
    
			int oilMass = 0;
			int  j = i;
			int count = 0;//用来计数,是否行驶了一周
			//寻找 i 最大能到达的站 y 
			while (oilMass >= 0 && count < n) {
    
    //判断当前位置能否行驶一周
				oilMass = oilMass + gas[j] - cost[j];
				j = (j + 1) % n;
				count++;
				if (oilMass >= 0 && count == n) {
    
    //如果行驶了一周,返回i
					return i;
				}
			}
			
			i = i + count;//如果不能行驶一周,i 直接移动到 y 的下一站
		}
		return -1;//不能行驶一周
    }
}

java code

Use, as shown in the figure below, to analyze the

main function of the source code: find the lowest point of the black line in the figure; if the sum is >=0, the starting point is the next position of the lowest point; if the sum is <0, there is no solution

class Solution {
    
    
    public int canCompleteCircuit(int[] gas, int[] cost) {
    
    
		int n = gas.length;
		int totalGas = 0;//总和
		int minGas = Integer.MAX_VALUE;
		int mindx = 0;
		for (int i = 0; i < n; i++) {
    
    
			totalGas += gas[i] - cost[i];
			if (totalGas < minGas) {
    
    //寻找最低点,记录下标
				minGas = totalGas;
				mindx = i;
			}
		}
		return totalGas >= 0 ? (mindx + 1) % n : -1;
    }
}