https://leetcode.com/problems/delete-and-earn/description/
跟买卖股票的还是一样的思路。
参考率 https://leetcode.com/problems/delete-and-earn/discuss/123634/C++-9ms-DP-solution
class Solution { public: int deleteAndEarn(vector<int>& nums) { int n = nums.size(); if (n == 0) return 0; vector< vector<int> > dp( nums.size() + 1, vector<int>(2) ); sort(nums.begin(), nums.end()); // 0表示不要这个数,1表示要这个数 dp[0][0] = 0; dp[0][1] = nums[0]; for (int i = 1; i < nums.size(); i++) { if (nums[i] == nums[i-1]) { dp[i][0] = dp[i-1][0]; dp[i][1] = dp[i-1][1] + nums[i]; continue; } dp[i][0] = max(dp[i-1][0], dp[i-1][1]); dp[i][1] = nums[i] == nums[i-1] + 1 ? dp[i-1][0] + nums[i] : dp[i][0] + nums[i]; } return max(dp[n-1][0], dp[n-1][1]); } };
https://leetcode.com/problems/delete-and-earn/discuss/109871/Awesome-Python-4-liner-with-explanation-Reduce-to-House-Robbers-Question
这里有个非常好的 hash转化成house robber的做法。当时我想到这个题感觉跟不能取相邻的XXX的感觉一样,但是没想到咋做。。
class Solution { public: int deleteAndEarn(vector<int>& nums) { vector<int> dp(10000+1, 0); for (int i = 0; i < nums.size(); i++) { dp[ nums[i] ] += nums[i]; } int last = 0, cur = 0; for (int i = 0; i < dp.size(); i++) { // cur其实是选了i-1的,last是cur之前的选择 int tmp = cur; cur = max(last + dp[i], cur); last = tmp; } return cur; } };
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