LintCode: Minimum Path Sum
C++,
time: O(m*n)
space: O(m*n)
1 class Solution { 2 public: 3 /** 4 * @param grid: a list of lists of integers. 5 * @return: An integer, minimizes the sum of all numbers along its path 6 */ 7 int minPathSum(vector<vector<int> > &grid) { 8 // write your code here 9 int m = grid.size(), n = grid[0].size(); 10 vector<vector<int> > dp(m, vector<int>(n)); 11 for (int i = 0; i < m; i++) { 12 for (int j = 0; j < n; j++) { 13 if (i == 0) { 14 if (j == 0) { 15 dp[i][j] = grid[i][j]; 16 } else { 17 dp[i][j] = dp[i][j-1] + grid[i][j]; 18 } 19 } else if (j == 0) { 20 dp[i][j] = dp[i-1][j] + grid[i][j]; 21 } else { 22 dp[i][j] = min(dp[i - 1][j], dp[i][j - 1]) + grid[i][j]; 23 } 24 } 25 } 26 return dp[m-1][n-1]; 27 } 28 };
C++,
time: O(m*n)
space: O(m)
1 class Solution { 2 public: 3 /** 4 * @param grid: a list of lists of integers. 5 * @return: An integer, minimizes the sum of all numbers along its path 6 */ 7 int minPathSum(vector<vector<int> > &grid) { 8 // write your code here 9 int m = grid.size(), n = grid[0].size(); 10 vector<int> dp(n); 11 for (int i = 0; i < m; i++) { 12 for (int j = 0; j < n; j++) { 13 if (i == 0) { 14 if (j == 0) { 15 dp[j] = grid[i][j]; 16 } else { 17 dp[j] = dp[j-1] + grid[i][j]; 18 } 19 } else if (j == 0) { 20 dp[j] = dp[j] + grid[i][j]; 21 } else { 22 dp[j] = min(dp[j], dp[j - 1]) + grid[i][j]; 23 } 24 } 25 } 26 return dp[n-1]; 27 } 28 };
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作者:
ZH奶酪(张贺)
邮箱:
cheesezh@qq.com
出处:
http://www.cnblogs.com/CheeseZH/
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