Lintcode: Unique Paths

C++

dp

递推式：dp[i][j] = dp[i-1][j] + dp[i][j-1]

初值：dp[i][j] = 1，i=0 or j=0

空间优化：省掉一维

class Solution {
public:
    /**
     * @param n, m: positive integer (1 <= n ,m <= 100)
     * @return an integer
     */
    int uniquePaths(int m, int n) {
        // wirte your code here
        vector<vector<int> > dp(m,vector<int>(n));
        for (int i = 0; i < m ; ++i) {
            for (int j = 0; j < n; ++j) {
                if ( i == 0 ) {
                    dp[i][j] = 1;
                } else if ( j == 0 ) {
                    dp[i][j] = 1;
                } else {
                    dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
                }
            }
        }
        return dp[m - 1][n - 1];
    }
};

空间优化

class Solution {
public:
    /**
     * @param n, m: positive integer (1 <= n ,m <= 100)
     * @return an integer
     */
    int uniquePaths(int m, int n) {
        // wirte your code here
        // vector<vector<int> > dp(m,vector<int>(n));
        vector<int> dp(n);
        for (int i = 0; i < m ; ++i) {
            for (int j = 0; j < n; ++j) {
                if ( i == 0 ) {
                    dp[j] = 1;
                } else if ( j == 0 ) {
                    dp[j] = 1;
                } else {
                    dp[j] = dp[j] + dp[j - 1];
                }
            }
        }
        return dp[n - 1];
    }
};