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Time: March 1 ~ March 13

  • Force buckle (LeetCode)- Sum of two numbers, valid parentheses, sum of two numbers | brush questions punch card – March 1
  • LeetCode – Merges two ordered lists, removes duplicates in sorted arrays,JavaScript Notes | Flash cards – March 2
  • Force Button (LeetCode)- Maximum suborder sum,JavaScript Data Structures and Algorithms (Arrays) | Brush questions punched -3 March
  • Say something about CSS | Tech Review – March 4
  • Force Buckle (LeetCode)- Stack, parenthesis generation | brush questions punch card – 5 March
  • It wasn’t that hard! Vue mall development | technical review – on March 6
  • Force buckle (LeetCode)- Plus one, queue | brush questions punch – 7 March
  • JavaScript data structure of the chain table | technical review – March 8
  • JavaScript Data Structures – Collections | Technical Review – March 9
  • LeetCode – Merge two ordered arrays, dictionary, hash table | swipe card – March 10
  • Force Buckle (LeetCode)- Symmetric binary Tree, Tree | Brush title punch card – March 11
  • Force buckle (LeetCode)-104. Maximum depth of binary tree, figure | brush problem clocking – March 12

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Stack, queue, linked list, set, dictionary and hash table, tree, graph

13. Convert Roman numerals to whole numbers

1. Title Description

Roman numerals contain the following seven characters: I, V, X, L, C, D and M.

Second, train of thought analysis

  1. Roman numerals are made up ofI,V,X,L,C,D,MA;
  2. When a small value is to the left of a large value, the value decreases, e.gIV=5-1=4
  3. When the small value is to the right of the large value, the small value is added, as inVI=5+1=6
  4. The rvalue is always positive, so the last digit must be positive

Putting a small value to the left of a large value is subtraction, otherwise addition.

  • All combinatorial possibilities are first listed and added to the hash table
  • And then I iterate over the string, because there are only two combinations, one is1Two characters, one is2A character, where2A character takes precedence over1A character
  • Determine if the combination of two characters exists in the hash table, and then extract the value and add it to the resultansMedium, and move back2A character. If it does not exist, the current is judged1If the character exists, the value is extracted and added to the resultansMedium, and move back1A character
  • The result is returned at the end of traversalans

Answer code

var romanToInt = function(s) { let map = { I: 1, IV: 4, V: 5, IX: 9, X: 10, XL: 40, L: 50, XC: 90, C: 100, CD: 400, D: 500, CM: 900, M: 1000 } let ans = 0; for(let i=0; i<s.length;) {if (I + 1 < s.l ength && map [s.s ubstring (I, I + 2)]) {/ / a group ans + 2 = map [s.s ubstring (I, I + 2)]; i += 2; }else{// single ans += map[s.substring(I, I +1)]; i++; } } return ans; };Copy the code

Four,

  1. Roman numerals to whole numbers

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