leetcode题目：Word Pattern （Javascript版）

本文链接：https://blog.youkuaiyun.com/Candy_home/article/details/48975103

本文介绍了一种用于检查字符串是否遵循给定模式的算法。通过多个示例解释了该算法的工作原理，并提供了一段详细的JavaScript代码实现，展示了如何比较模式与字符串的关系。

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Given a pattern and a string str, find if str follows the same pattern.

Examples:

pattern = "abba", str = "dog cat cat dog" should return true.
pattern = "abba", str = "dog cat cat fish" should return false.
pattern = "aaaa", str = "dog cat cat dog" should return false.
pattern = "abba", str = "dog dog dog dog" should return false.

Notes:

patterncontains only lowercase alphabetical letters, and str contains words separated by a single space. Each word in str contains only lowercase alphabetical letters.
Both pattern and str do not have leading or trailing spaces.
Each letter in pattern must map to a word with length that is at least 1.

答案：

/**
 * @param {string} pattern
 * @param {string} str
 * @return {boolean}
 */

var wordPattern = function(pattern, str) {

//返回结果
var result = true;

//分割数组
var patternArray = pattern.split("");
var strArray = str.split(" ");

if (patternArray.length != strArray.length){
  result = false;
}

//数组之间的关系
var relationship = [];
var relationshipNegative = [];

//如果相等，就放入relationship数组中
//如果不相等，就放入relationshipNegative数组中
for (var i = 0; i < patternArray.length; i++) {
  for (var j = 1; j < patternArray.length; j++) {
    if ( i!==j ) {
      if (patternArray[i] === patternArray[j]) {
        relationship.push({i:i,j:j});
      }
      else{
        relationshipNegative.push({i:i,j:j});
      }
    }
  }
}

//判断是否符合相等关系的描述，不符合返回false
for (var k = 0; k < relationship.length; k++) {
  // console.warn(strArray[relationship[k].i] !== strArray[relationship[k].j]);
  if (strArray[relationship[k].i] !== strArray[relationship[k].j] ) {
    // console.log("become false");
    result = false;
  }
}

//判断是否符合不相等关系的描述，不符合返回false
for (var h = 0; h < relationshipNegative.length; h++) {
  if (strArray[relationshipNegative[h].i] === strArray[relationshipNegative[h].j] ) {
    result = false;
  }
}

return result;
};