思路：两个指针，从零开始，比较两个数组里面的指针，每次把两者之中最小的放到结果数组中来。
const getOrderList = (list1, list2) => { let newList = []; let i = 0; let j = 0; while (i < list1.length && j < list2.length) { if (list1[i] <= list2[j]) { newList.push(list1[i]); i++; } else { newList.push(list2[j]); j++; } } if (i === list1.length) { newList = newList.concat(list2.slice(j)); } if (j === list2.length) { newList = newList.concat(list1.slice(i)); } return newList; };
2、判断一个单词是否回文
思路：把字符串翻转之后，看是否等于原来的字符
const isRecerseValue = (value = ‘‘) => { let newValue = ‘‘ let i = value.length-1 while(i>=0) { newValue +=value[i] i--; } return newValue === value } console.log(isRecerseValue(‘aba‘))
原文：https://www.cnblogs.com/jwenming/p/14702670.html