leetcode 刷题记

1. Construct Binary Tree from Preorder and Inorder Traversal
   思路：1.先序第一个值简历root 2.找到左子树和右子树，递归构建
   问题：内存溢出
2. 64 Minimum Path Sum
   思路： dp[i][j] = min(dp[i-1][j], dp[i][j-1]) + grid[i][j]) (i > 0,j > 0)
   dp[i][j] = dp[i][j-1] + grid[i][j]) (i == 0)
   dp[i][j] = dp[i-1][j] + grid[i][j]) (j == 0)
   注意：条件的初始化和递归公式的精确性表达
   二维vector 初始化：

vector<vector<int> > ways(m, vector<int> (n, 1));

int minPathSum(vector<vector<int>>& grid) {
        if (grid.size() == 0) {
            return 0;
        }
        int rows = grid.size();
        int cols = grid[0].size();
        vector<vector<int>> sumgrid;
        for (int i = 0; i < rows; i++) {
            vector<int> row(cols, 0);
            sumgrid.push_back(row);
        }
        for (int i = 0; i < rows; i++) {
            for (int j = 0; j < cols; j++) {
                if (i == 0 && j == 0) {
                    sumgrid[i][j] = grid[i][j];
                } else if (i == 0) {
                    sumgrid[i][j] = sumgrid[i][j - 1] + grid[i][j];
                } else if (j == 0) {
                    sumgrid[i][j] = sumgrid[i- 1][j] + grid[i][j];
                } else {
                    sumgrid[i][j] = min(sumgrid[i - 1][j], sumgrid[i][j - 1]) + grid[i][j];
                }
            }
        }
        return sumgrid[rows - 1][cols - 1];
     }

1. 216 Combination Sum III
   思路：回溯

void combination(vector<vector<int>> &result, vector<int> ans, int k, int n) {
        if (ans.size() == k && n == 0) {
            result.push_back(ans);
            **return;**
        }
        if (ans.size() < k)  {
            for (int i = ans.empty() ? 1 : ans.back() + 1; i <= 9; i++) {
                if (n - i < 0) {
                    break;
                }
                **ans.push_back(i);
                combination(result, ans, k, n - i);
                ans.pop_back();**
            }
        }
    }
    vector<vector<int>> combinationSum3(int k, int n) {
        vector<vector<int>> result;
        vector<int> ans;
        combination(result, ans, k, n);
        return result;
    }