更新毎日解题06月04日 #24
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更新毎日解题06月04日 #24
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azl397985856
requested changes
Jun 10, 2019
templates/daily/2019-06-04.md
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| Return the starting gas station's index if you can travel around the circuit once in the clockwise direction, otherwise return -1. | ||
| ``` | ||
| ## 参考答案 | ||
| `1.暴力求解，时间复杂度O(n^2) |
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改成：
1.暴力求解，时间复杂度O(n^2)
我们可以一次遍历gas，对于每一个gas我们依次遍历后面的gas，计算remian，如果remain一旦小于0，就说明不行，我们继续遍历下一个
这种格式
templates/daily/2019-06-04.md
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| retirn -1; | ||
| ``` | ||
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| `2.比较巧妙的方法，时间复杂度是O(n) |
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| `2.比较巧妙的方法，时间复杂度是O(n) | ||
| 这个方法基于两点： | ||
| 2-1:如果站点i到达站点j走不通,那么从i到j之间的站点（比如k）出发一定都走不通。前提i（以及i到k之间）不会拖累总体（即remain >= 0）。 |
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改成：
这个方法基于两点：
(注意这里有一个空行)
2-1:如果站点i到达站点j走不通,那么从i到j之间的站点（比如k）出发一定都走不通。前提i（以及i到k之间）不会拖累总体（即remain >= 0）。
templates/daily/2019-06-04.md
Outdated
| `2.比较巧妙的方法，时间复杂度是O(n) | ||
| 这个方法基于两点： | ||
| 2-1:如果站点i到达站点j走不通,那么从i到j之间的站点（比如k）出发一定都走不通。前提i（以及i到k之间）不会拖累总体（即remain >= 0）。 | ||
| 2-2:如果cost总和大于gas总和，无论如何也无法走到终点，这个比较好理解。因此假如存在一个站点出发能够到达终点，其实就说明cost总和一定小于等于gas总和` |
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