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Difficulty level: medium
1. Description
Given a m x ngrid grid, find a path from the upper left corner to the lower right corner such that the sum of the numbers on the path is the smallest.
Description: You can only move down or to the right one step at a time.
2. Example
Example 1
输入：grid = [[1,3,1],[1,5,1],[4,2,1]]
输出：7
解释：因为路径 1→3→1→1→1 的总和最小。
复制代码Example 2
输入：grid = [[1,2,3],[4,5,6]]
输出：12
复制代码Restrictions:
m == grid.lengthn == grid[i].length1 <= m, n <= 2000 <= grid[i][j] <= 100
3. Answers
class MinimumPathSum {
func minPathSum(_ grid: [[Int]]) -> Int {
guard grid.count != 0 && grid[0].count != 0 else{
return 0
}
let m = grid.count, n = grid[0].count
var dp = Array(repeating: Array(repeating: 0, count: n), count: m)
for i in 0..<m {
for j in 0..<n {
if i == 0 && j == 0{
dp[i][j] = grid[i][j]
} else if i == 0 {
dp[i][j] = dp[i][j - 1] + grid[i][j]
} else if j == 0 {
dp[i][j] = dp[i - 1][j] + grid[i][j]
} else {
dp[i][j] = min(dp[i][j - 1], dp[i - 1][j]) + grid[i][j]
}
}
}
return dp[m - 1][n - 1]
}
}
复制代码- Main idea: Classical 2D dynamic programming.
- Time Complexity: O(mn)
- Space Complexity: O(mn)
Repository for the algorithm solution: LeetCode-Swift
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