674. Longest Continuous Increasing Subsequence
2017-12-26 17:34
330 查看
看了别人的一句 int max=1(不断更新max。。 思路!!!!)后 写了个简单的,不需要vector保存结果。
class Solution {
public:
int findLengthOfLCIS(vector<int>& nums) {
int a=nums.size();
int cnt=1;
if (a==0) return 0;
int max=1;
if (a>=2){
for(int i=1;i<a;i++)
{
if( nums[i]<=nums[i-1])
{
if (max<cnt) max=cnt;
cnt=1;
}
else if(i==a-1&&nums[i]>nums[i-1])
{
cnt=cnt+1;
if (max<cnt) max=cnt;
}
else
cnt=cnt+1;
}
}
return max;
}
};
其实可以更简洁:(应为if可以不要,a=1时i<a不满足,跳过for 循环 max=1不变)
class Solution {
public:
int findLengthOfLCIS(vector<int>& nums) {
int a=nums.size();
int cnt=1;
if (a==0) return 0;
int max=1;
for(int i=1;i<a;i++)
{
if( nums[i]<=nums[i-1])
{
if (max<cnt) max=cnt;
cnt=1;
}
else if(i==a-1&&nums[i]>nums[i-1])
{
cnt=cnt+1;
if (max<cnt) max=cnt;
}
else
cnt=cnt+1;
}
return max;
}
};
class Solution {
public:
int findLengthOfLCIS(vector<int>& nums) {
int a=nums.size();
int cnt=1;
if (a==0) return 0;
int max=1;
if (a>=2){
for(int i=1;i<a;i++)
{
if( nums[i]<=nums[i-1])
{
if (max<cnt) max=cnt;
cnt=1;
}
else if(i==a-1&&nums[i]>nums[i-1])
{
cnt=cnt+1;
if (max<cnt) max=cnt;
}
else
cnt=cnt+1;
}
}
return max;
}
};
其实可以更简洁:(应为if可以不要,a=1时i<a不满足,跳过for 循环 max=1不变)
class Solution {
public:
int findLengthOfLCIS(vector<int>& nums) {
int a=nums.size();
int cnt=1;
if (a==0) return 0;
int max=1;
for(int i=1;i<a;i++)
{
if( nums[i]<=nums[i-1])
{
if (max<cnt) max=cnt;
cnt=1;
}
else if(i==a-1&&nums[i]>nums[i-1])
{
cnt=cnt+1;
if (max<cnt) max=cnt;
}
else
cnt=cnt+1;
}
return max;
}
};
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