There are N gas stations along a circular route, where the amount of gas at station i is gas[i].
You have a car with an unlimited gas tank and it costs cost[i] of gas to travel from station i to its next station (i+1). You begin the journey with an empty tank at one of the gas stations.
Return the starting gas station's index if you can travel around the circuit once, otherwise return -1.
Note:The solution is guaranteed to be unique.
遍歷的方法，是對(duì)的，但是超時(shí)了：

var canCompleteCircuit = function(gas, cost) {
    var start = 0;
    var num = gas.length;
    while (start<gas.length) {
        var remain = 0;
        for (let i = start; i < num; i++) {
            remain += gas[i] - cost[i];
            if (remain<=0)
                break;
        }
        if (remain>=0&&start===0) {
            return start;
        }
        if (remain<=0) {
            start++;
            continue;
        }
        for (var i = 0; i < start; i++) {
            remain += gas[i] - cost[i];
            if (remain<=0)
                break;
        }
        if (remain>=0&&i === start - 1)
            return start;
        start++;
    }
    return -1;
};

仔細(xì)想想，有兩點(diǎn)：
如果油的總量大于需要花費(fèi)的總量，一定存在解。
如果一輛車(chē)從A開(kāi)始，到達(dá)不了B，那么從A和B之間任何一站開(kāi)始都到達(dá)不了B。
所以我們從頭開(kāi)始遍歷數(shù)組，并記錄油箱中的剩余，如果小于0了，那就意味著此次出發(fā)的點(diǎn)到不了當(dāng)前點(diǎn)，得從當(dāng)前點(diǎn)的下一點(diǎn)出發(fā)。遍歷到最后，判定一下是否有唯一解，返回最后設(shè)定的起點(diǎn)。

var canCompleteCircuit = function(gas, cost) {
    var start = 0;
    var total = 0;
    var tank = 0;
    //if car fails at 'start', record the next station
    for(var i = 0;i < gas.length;i++) {
        tank = tank + gas[i] - cost[i];
        if(tank < 0) {
            start = i + 1;
            total += tank;
            tank = 0;
        }
    }
    return (total + tank < 0) ? -1 : start;
};