1、题目描述

2、题目分析

使用动态规划,在计算以每个字符结尾的最长子序列。

3、代码

 int lengthOfLIS(vector<int>& nums) {
if(nums.size() == ){
return ;
} vector<int> dp(nums.size() , );
int maxans = ;
for(int i = ; i < nums.size(); i++){
int maxval = ;
for( int j = ; j < i; j++){
if(nums[i] > nums[j]){
maxval = max(dp[j], maxval);
}
}
dp[i] = maxval + ;
maxans = max(dp[i], maxans);
} return maxans;
}

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