leetcode 动态规划类型题

Posted 2021-02-17 dreamcatcher-zp

1，Triangle

1 int mininumTotal(vector<vector<int>>& triangle)  {
2     for (int i = triangle.size() - 2; i >= 0; --i) {  
3         for (int j = 0; j < i + 1; ++j) {
4             // 从下往上依次保存当前路径的最小值，上层只会用到下层的最小值
5             triangle[i][j] += min(triangle[i + 1][j], triangle[i + 1][j + 1]);
6         }
7     }
8     return triangle[0][0];
9 }

triangle

 2，Maximum SubArray

 1 /*
 2  * 状态转移方程为：f[j] = max{ f[j-1] + S[j],S[j] },其中  1 <= j <= n
 3  *                 target = max{ f[j] },其中  1 <= j <= n
 4 */
 5 int maxArray(vector<int>& nums) {
 6     int n = nums.size();
 7     vector<int> dp(n + 1);
 8     dp[0] = 0;
 9     for (int i = 0; i < n; ++i) {
10         dp[i + 1] = max(dp[i] + nums[i], nums[i]);
11     }
12     return *max_element(dp.begin(), dp.end());

maxArray

3，