摘要：
https://leetcode-cn.com/problems/fruit-into-baskets/暴力法则：timeout functotalFruit（tree[]int）int｛res:=0n:=len（tree）i，j:=0,0fori=0；i˂n；i++{pre1，pre2:=-1，-1forj=i；j˂n；j++{ifpre1==-1||pre1==tree[j]{pre1=tree[j]con
https://leetcode-cn.com/problems/fruit-into-baskets/
暴力法：超时
func totalFruit(tree []int) int {
res := 0
n := len(tree)
i,j := 0,0
for i=0;i<n;i++{
pre1,pre2 := -1,-1
for j=i;j<n;j++{
if pre1 == -1 || pre1 == tree[j]{
pre1 = tree[j]
continue
}else if pre2 == -1 || pre2 == tree[j]{
pre2 = tree[j]
continue
}else{
if tree[j] == pre1 || tree[j] == pre2{
continue
}
res = MAX(res,j-i)
break
}
}
res = MAX(res,j-i)
}
return res
}
func MAX(i,j int) int{
if i<j{
return j
}else{
return i
}
}
用滑动窗口优化下时间复杂度：最主要一点，
tRight := right-1 pre1 = tree[tRight]
pre2 = tree[right]
func totalFruit(tree []int) int {
res := 0
n := len(tree)
left,right := 0,0
pre1,pre2 := -1,-1
for right < n{
if pre1 == -1 || pre1 == tree[right]{
pre1 = tree[right]
right++
continue
}else if pre2 == -1 || pre2 == tree[right]{
pre2 = tree[right]
right++
continue
}else{
res = MAX(res,right-left)
tRight := right-1
pre1 = tree[tRight]
pre2 = tree[right]
for tree[tRight] == pre1{
tRight--
}
left = tRight+1
}
}
res = MAX(res,right-left)
return res
}
func MAX(i,j int) int{
if i<j{
return j
}else{
return i
}
}