javascript - 根据另一个数组修改一个数组

Question

我有一个这样的短语引用数组:

const reference = ["your", "majesty", "they", "are", "ready"];

我想基于另一个数组加入上述数组的一些元素，所以如果 another这是:

const another = ["your", "they are"];

结果将是这样的:

result = ["your", "majesty", "they are", "ready"];

另一种情况要更清楚:

const reference = ["your", "majesty", "they", "are", "ready"];
const another = ["your majesty", "they are ready"];
result = ["your majesty", "they are ready"];

或者:

const reference = ["your", "majesty", "they", "are", "ready"];
const another = ["majesty they", "ready"];
result = ["your", "majesty they", "are", "ready"];

当然如果有一个空的 another我们得到与结果相同的引用:

const reference = ["your", "majesty", "they", "are", "ready"];
const another = [];
result = ["your", "majesty", "they", "are", "ready"];

Answer 1

const reference = ["your", "majesty", "they", "are", "ready"];
//const another = ["your", "they are"];
//const another = ["your majesty", "they are ready"];
//const another = ["majesty they", "ready"];
const another = [];

function formatResult(reference, another) {
  //1: another is [] then return reference
  if (another.length == 0) {
    return reference;
  } else {
    //2: format the reference array with another array
    another.forEach(x => {
      var words = x.split(' ');
      if (words.length > 1) {
        var index = 0;
        var sindex = 0; // if the index are countiues
        for (var i = 0; i < words.length; i++) {
          //find this word in refference array.
          index = reference.indexOf(words[i]);
          if (i == 0) {
            sindex = index;
          } else {
            //check if the index are next to each other, if not break the loop
            if (sindex != index) {
              break;
            }
          }
          sindex++;
        }
        if (index == sindex - 1) //to check all the index in the order
        {
          var startIndex = index - (words.length - 1);
          //update reference array with another array.
          reference.splice(startIndex, words.length, x);
        }
      }
    })
  }
  return reference;
}
var result = formatResult(reference, another);
console.log(result);

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