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preface

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Difficulty level: Difficult

1. Describe

Significant numbers (in order) can be broken down into the following parts:

  1. A decimal or whole number
  2. (Optional) One'e' 或 'E', followed by an integer

Decimals (in order) can be divided into the following parts:

  1. (Optional) A symbolic character ('+' 或 The '-')
  2. One of the following formats:
    1. At least one digit followed by a dot'. '
    2. At least one digit followed by a dot'. ', followed by at least one digit
    3. A point'. ', followed by at least one digit

Integers (in order) can be divided into the following parts:

  1. (Optional) A symbolic character ('+' 或 The '-')
  2. At least one digit

Some valid number listed below: [” 2 “, “0089”, “0.1”, “+ 3.14”, “4”, “- 9”, “2 e10,” “- 90 e3”, “3 e + 7”, “+ 6 e – 1”, “53.5 e93”, “123.456 e789”]

Invalid part number listed below: [” ABC “, “1 a”, “1 e”, “e3”, “99 e2. 5”, “- 6”, “- + 3”, “95 a54e53”]

Given the string s, return true if s is a valid number.

Example 2.

Example 1

Input: s = 0 Output: trueCopy the code

Example 2

Input: s = "e" Output: falseCopy the code

Example 3

Input: s = "." Output: falseCopy the code

Constraints:

  • 1 <= s.length <= 20
  • sContains only English letters (uppercase and lowercase), digits (0-9), plus'+'A minus sign,The '-', or some'. ' 。

3. The answer

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]}}Copy the code
  • Main idea: Iterate through the string and process the point, exponent, number, and symbol use cases separately.
  • Time complexity: O(n)
  • Space complexity: O(1)

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