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Commit1015a0c

Browse files
committed
add new files
1 parent3cad8d8 commit1015a0c

6 files changed

+127
-5
lines changed

‎122. Best Time to Buy and Sell Stock II.go

Lines changed: 8 additions & 2 deletions
Original file line numberDiff line numberDiff line change
@@ -5,8 +5,14 @@ import "math"
55
funcmaxProfit(prices []int)int {
66
cur,max:=0,0
77
fori:=1;i<len(prices);i++ {
8-
cur=int(math.Max(float64(cur),float64(cur+prices[i]-prices[i-1])))
9-
max=int(math.Max(float64(cur),float64(max)))
8+
cur=mymax(cur,cur+prices[i]-prices[i-1])
9+
max=mymax(cur,max)
1010
}
1111
returnmax
1212
}
13+
funcmymax(x,yint)int {
14+
ifx>y {
15+
returnx
16+
}
17+
returny
18+
}

‎125. Valid Palindrome.go

Lines changed: 1 addition & 3 deletions
Original file line numberDiff line numberDiff line change
@@ -1,19 +1,17 @@
11
package main
22

33
import (
4-
"fmt"
54
"regexp"
65
"strings"
76
)
87

98
//TODO 速度不是很理想
10-
funcisPalindrome1(sstring)bool {
9+
funcisPalindrome(sstring)bool {
1110
pat:="[,:.@#--?\";!` ]"
1211
re,_:=regexp.Compile(pat)
1312

1413
s=re.ReplaceAllString(s,"")
1514
s=strings.ToLower(s)
16-
fmt.Println("s=",s)
1715
ifs=="" {
1816
returntrue
1917
}

‎134. Gas Station.go

Lines changed: 24 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,24 @@
1+
package main
2+
3+
/**
4+
如果一个数组的总和非负，那么一定可以找到一个起始位置，从他开始绕数组一圈，累加和一直都是非负的
5+
1. 假如从位置i开始，i+1，i+2...，一路开过来一路油箱都没有空。说明什么？说明从i到i+1，i+2，...肯定是正积累。
6+
2. 现在突然发现开往位置j时油箱空了。这说明什么？说明从位置i开始没法走完全程(废话)。那么，我们要从位置i+1开始重新尝试吗？不需要！为什么？因为前面已经知道，位置i肯定是正积累，那么，如果从位置i+1开始走更加没法走完全程了，因为没有位置i的正积累了。同理，也不用从i+2，i+3，...开始尝试。所以我们可以放心地从位置j+1开始尝试。
7+
TODO 然而并不是AC里面最快的
8+
*/
9+
funccanCompleteCircuit(gas []int,cost []int)int {
10+
start,remain,debt:=0,0,0
11+
fori:=0;i<len(gas);i++ {
12+
13+
remain+=gas[i]-cost[i]
14+
ifremain<0 {
15+
debt+=remain
16+
start=i+1
17+
remain=0
18+
}
19+
}
20+
ifremain+debt>=0 {
21+
returnstart
22+
}
23+
return-1
24+
}

‎137. Single Number II.go

Lines changed: 25 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,25 @@
1+
package main
2+
3+
/**
4+
这是一个简单的思路
5+
貌似可以用位运算搞定。。。
6+
*/
7+
funcsingleNumber(nums []int)int {
8+
k:=make(map[int]int)
9+
// count=0
10+
for_,kvs:=rangenums {
11+
if_,ok:=k[kvs];ok {
12+
k[kvs]+=1
13+
14+
}else {
15+
k[kvs]=1
16+
}
17+
}
18+
p:=0
19+
forl:=rangek {
20+
ifk[l]==1 {
21+
p=l
22+
}
23+
}
24+
returnp
25+
}

‎152. Maximum Product Subarray.go

Lines changed: 32 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,32 @@
1+
package main
2+
3+
/**
4+
考虑到负负得正的情况，需要同时记录最大值和最小值
5+
*/
6+
funcmaxProduct(nums []int)int {
7+
prevMax,prevMin,ret:=nums[0],nums[0],nums[0]
8+
fori:=1;i<len(nums);i++ {
9+
ifnums[i-1]==0 {
10+
prevMax,prevMin=nums[i],nums[i]
11+
}else {
12+
curMax:=max(nums[i],max(nums[i]*prevMax,nums[i]*prevMin))
13+
curMin:=min(nums[i],min(nums[i]*prevMax,nums[i]*prevMin))
14+
prevMax,prevMin=curMax,curMin
15+
}
16+
ret=max(prevMax,ret)
17+
}
18+
returnret
19+
}
20+
21+
funcmax(x,yint)int {
22+
ifx>y {
23+
returnx
24+
}
25+
returny
26+
}
27+
funcmin(x,yint)int {
28+
ifx>y {
29+
returny
30+
}
31+
returnx
32+
}

‎165. Compare Version Numbers.go

Lines changed: 37 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,37 @@
1+
package main
2+
3+
import (
4+
"strconv"
5+
"strings"
6+
)
7+
8+
/**
9+
首先切分字符串，然后循环比较每一段，注意处理连续0的情况
10+
*/
11+
12+
funccompareVersion(version1string,version2string)int {
13+
v1:=strings.Split(version1,".")
14+
v2:=strings.Split(version2,".")
15+
ret:=1
16+
i:=0
17+
fori=0;i<len(v1)&&i<len(v2);i++ {
18+
a,_:=strconv.Atoi(v1[i])
19+
b,_:=strconv.Atoi(v2[i])
20+
ifa>b {
21+
return1
22+
}elseifa<b {
23+
return-1
24+
}
25+
}
26+
iflen(v1)<len(v2) {
27+
ret=-1
28+
v1=v2
29+
}
30+
for ;i<len(v1);i++ {
31+
a,_:=strconv.Atoi(v1[i])
32+
ifa!=0 {
33+
returnret
34+
}
35+
}
36+
return0
37+
}

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