使用 for 循环或 switch 语句的 JavaScript 序列

Question

我遇到了一个问题以及如何解决它。请注意，我是 JavaScript 的新手，我觉得这个问题过于复杂了。

问题:

Given a sequence of integers as an array, determine whether it is possible to obtain a strictly increasing sequence by removing no more than one element from the array.

例子

对于 sequence = [1, 3, 2, 1]，输出应该是 almostIncreasingSequence(sequence) = false;

为了得到严格递增的序列，这个数组中没有一个元素可以被删除。

对于 sequence = [1, 3, 2]，输出应该是 almostIncreasingSequence(sequence) = true。

您可以从数组中删除 3 以获得严格递增序列 [1, 2]。或者，您可以删除 2 以获得严格递增序列 [1, 3]。

感谢您的所有评论!我想将其更新为一个更好的问题，并通知您我已经找到了解决方案，如果有人想检查它并查看是否有更简洁的方式来放置它。 :)

function almostIncreasingSequence(sequence) {    
  if(sequence.length == 2) return true;
  var error = 0;
  for(var i = 0; i < sequence.length - 1; i++){
    if(sequence[i] >= sequence[i+1]){
      var noStepBack = sequence[i-1] && sequence[i-1] >= sequence[i+1];
      var noStepFoward = sequence[i+2] && sequence[i] >= sequence[i+2];
      if(i > 0 && noStepBack && noStepFoward) {
        error+=2;
      }else{
        error++;
      }
    }
    if(error > 1){
      return false;
    }
  }
  return true;
}

Answer 1

想想你的代码:

sequence[i+1] - sequence[i] !== 1, variable++;

适用于以下数组:[1,2,3,8,8]。

从问题描述来看，不清楚程序是否必须删除一个字符。但如果是这样的话，下面的代码应该可以做到。

function canGetStrictlyIncreasingSeq(numbers) {
  var counter = 0;
  var lastGreatestNumber = numbers[0];
  for (var i = 1; i < numbers.length; i++) {
    if (lastGreatestNumber >= numbers[i]) {
      counter++;
      lastGreatestNumber = numbers[i];
    } else {
      lastGreatestNumber = numbers[i];
    }
  }
  if (counter <= 1)
    return true;
  return false;
}

var nums1 = [1, 2, 3, 4, 5]; //true
var nums2 = [1, 2, 2, 3, 4]; //true
var nums3 = [1, 3, 8, 1, 9]; //true
var nums4 = [3, 2, 5, 6, 9]; //true
var nums5 = [3, 2, 1, 0, 5]; //false
var nums6 = [1, 2, 2, 2, 3]; //false
var nums7 = [1, 1, 1, 1, 1]; //false
var nums8 = [1, 2]; //true
var nums9 = [1, 2, 2]; //true
var nums10 = [1, 1, 2, 3, 4, 5, 5]; //false
var nums11 = [10, 1, 2, 3, 4, 5]; //true
var nums12 = [1, 2, 3, 4, 99, 5, 6]; //true


console.log(canGetStrictlyIncreasingSeq(nums1));
console.log(canGetStrictlyIncreasingSeq(nums2));
console.log(canGetStrictlyIncreasingSeq(nums3));
console.log(canGetStrictlyIncreasingSeq(nums4));
console.log(canGetStrictlyIncreasingSeq(nums5));
console.log(canGetStrictlyIncreasingSeq(nums6));
console.log(canGetStrictlyIncreasingSeq(nums7));
console.log(canGetStrictlyIncreasingSeq(nums8));
console.log(canGetStrictlyIncreasingSeq(nums9));
console.log(canGetStrictlyIncreasingSeq(nums10));
console.log(canGetStrictlyIncreasingSeq(nums11));
console.log(canGetStrictlyIncreasingSeq(nums12));