leedcode-1563. 石子游戏 V

目录

 

 思路一：（区间dp）

dp[i][j] 表示合并区间 i ~ j 所能获得的最大收益

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class Solution { public:       int a[505];     int dp[505][505];     int stoneGameV(vector<int>& stoneValue) {         int len = stoneValue.size();           for(int i = 1; i <= len; i++) {             a[i] = stoneValue[i-1];             a[i] += a[i-1];         }           for(int ls = 2; ls <= len; ls++){ // 长度             for(int i = 1; i <= len; i++) { // 左端                 int j = i + ls - 1; //右端                 if (j > len) break;                 for(int k = i; k < j; k++){                     int sum_l = a[k] - a[i-1];                     int sum_r = a[j] - a[k];                     if (sum_l > sum_r) dp[i][j] = max(dp[i][j], dp[k+1][j] + sum_r);                     else if (sum_l < sum_r) dp[i][j] = max(dp[i][j], dp[i][k] + sum_l);                     else dp[i][j] = max(dp[i][j], max(dp[i][k]+sum_l, dp[k+1][j]+sum_r));                 }             }         }         return dp[1][len];     } };

　　

思路二：（记忆化搜索）

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class Solution { public:     int sum[505];     int dp[505][505];     int dfs(int l, int r){         if (l >= r) return 0;         if (dp[l][r]) return dp[l][r];           int ans = 0;         for(int i = l; i < r; i++){             int s1 = sum[i]-sum[l-1];             int s2 = sum[r]-sum[i];             if(s1 < s2) ans = max(ans, s1 + dfs(l, i));             else if (s1 > s2) ans = max(ans, s2 + dfs(i+1, r));             else ans = max(ans, max(dfs(l, i), dfs(i+1, r))+s1);         }         return dp[l][r] = ans;     }       int stoneGameV(vector<int>& stoneValue) {         sum[1] = stoneValue[0];         for(int i = 1; i < stoneValue.size(); i++) {             sum[i+1] = sum[i] + stoneValue[i];         }           int len = stoneValue.size();           return dfs(1, len);     } };

　　

作者：静默虚空
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