Commit ed609a1

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Merge pull request SharingSource#61 from SharingSource/ac_oier

✨feat: Modify 1736 & 1893

2 parents 3ff438e + b3522a2 commit ed609a1Copy full SHA for ed609a1

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`@@ -53,7 +53,6 @@ Tag : 「贪心」`
`53`	`53`	`代码:`
`54`	`54`	```Java
`55`	`55`	`class Solution {`
`56`		`- static`
`57`	`56`	`public String maximumTime(String time) {`
`58`	`57`	`StringBuilder sb = new StringBuilder();`
`59`	`58`	`sb.append(time.charAt(0) == '?' ? (time.charAt(1) == '?' \|\| time.charAt(1) < '4') ? '2' : '1' : time.charAt(0));`

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`@@ -124,7 +124,7 @@ class Solution {`
`124`	`124`
`125`	`125`	`在朴素的「树状数组」解法中,我们无法直接查询 $[l, r]$ 区间中被覆盖过的个数的根本原因是「某个值可能会被重复添加到树状数组中」。`
`126`	`126`
`127`		`-因此,一种更加优秀的做法:在往树状数组中添树的时候进行去重,然后通过 $cnt = query(r) - query(l - 1)$ 直接得出 $[l, r]$ 范围内有多少个数被添加过。`
	`127`	`+因此,一种更加优秀的做法:在往树状数组中添数的时候进行去重,然后通过 $cnt = query(r) - query(l - 1)$ 直接得出 $[l, r]$ 范围内有多少个数被添加过。`
`128`	`128`
`129`	`129`	这样的 `Set` 去重操作可以使得我们查询的复杂度从 $O(n\log{C})$ 下降到 $O(\log{C})$。
`130`	`130`

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