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🐱(dynamic): 983. 最低票价

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Original file line number	Diff line number	Diff line change
`@@ -1508,6 +1508,35 @@ class Solution:`
`1508`	`1508`	`return len(res)`
`1509`	`1509`	```
`1510`	`1510`
	`1511`	`+## 983. 最低票价`
	`1512`	`+`
	`1513`	`+[原题链接](https://leetcode-cn.com/problems/minimum-cost-for-tickets/)`
	`1514`	`+`
	`1515`	`+### 解一:动态规划`
	`1516`	`+`
	`1517`	+从后往前进行动态规划,用 `dp[i]` 代表从第 `i` 天到最后一天需要的最低票价。
	`1518`	`+`
	`1519`	+- 当 `i` 无需出行时,依据贪心法则:`dp[i] = dp[i + 1]`,即在第 `i` 天无需购票
	`1520`	+- 当 `i` 需要出行,`dp[i] = min(costs[j] + dp[i + j])`,`j` 的取值是 1/7/30。如果第 `i` 天出行,我们买了 `j` 天的票,那么后续 `j` 天都不需要购票事宜了,所以只要加上 `dp[i + j]` 的票价即可。
	`1521`	`+`
	`1522`	+```python
	`1523`	`+class Solution:`
	`1524`	`+ def mincostTickets(self, days: List[int], costs: List[int]) -> int:`
	`1525`	`+ ans = 0`
	`1526`	`+ dp = [0 for _ in range(400)]`
	`1527`	`+ for i in range(365, 0, -1):`
	`1528`	`+ if i in days:`
	`1529`	`+ # 需要出行`
	`1530`	`+ min_dp = float('inf')`
	`1531`	`+ min_dp = min(costs[0] + dp[i + 1], costs[1] + dp[i + 7], costs[2] + dp[i + 30])`
	`1532`	`+ dp[i] = min_dp`
	`1533`	`+ else:`
	`1534`	`+ # 不需要出行`
	`1535`	`+ dp[i] = dp[i + 1]`
	`1536`	`+ # print(dp[20])`
	`1537`	`+ return dp[1]`
	`1538`	+```
	`1539`	`+`
`1511`	`1540`	`## 1137. 第 N 个泰波那契数`
`1512`	`1541`
`1513`	`1542`	`[原题链接](https://leetcode-cn.com/problems/n-th-tribonacci-number/)`

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