8.27 京东前端笔试
思路：就是分别处理一下，前面的toUpperCase()，后面的toLowerCase()；
let str = "ABCba"，n = 3; console.log(str.substring(0, n).toUpperCase() + str.substring(n).toLowerCase());
2. 打印1-n^2
思路：自己先写几个找找规律，n为奇数，直接从1-n^2顺序打印，每行输出n个；n为偶数，直接从1-n^2顺序打印，每行输出n个，但是奇偶行的顺序得相反，比如n=2时应该是[[1,2],[4,3]]，这个只要一行用push、一行用unshift就好了
function getArr(n) {
if(n % 2 === 1) {
let index = n, val = 1;
while(index) {
let arr = []
for(let i = 0; i < n; ++i) {
arr.push(val)
val++;
}
console.log(arr.join(' '));
index--
}
} else if(n % 2 === 0) {
let val = 1, index = n;
while (index) {
let arr = [];
for (let i = 0; i < n; ++i) {
if(index % 2 === 0) {
arr.push(val);
} else {
arr.unshift(val);
}
val++;
}
console.log(arr.join(" "));
index--;
}
}
}
getArr(4);
3. 长城数，8.20号网易也考了长城数，虽然不完全一样，但是思路有借鉴的地方。最终要保证奇、偶项分别相同，那么就用map记录下奇数项各个值的个数、偶数项各个值的个数，各自取个数最多的那个；注意，如果奇偶个数最多的数相同，就要考虑第二多的了，有一个要变为第二多的那个数，看操作谁的次数最少就用哪个。
function getTimes(arr) {
let oddMap = new Map(),
evenMap = new Map();
for(let i = 0; i < arr.length; i++) {
if(i % 2 === 0) {
oddMap.set(arr[i], (oddMap.get(arr[i]) || 0) + 1);
} else {
evenMap.set(arr[i], (evenMap.get(arr[i]) || 0) + 1);
}
}
// // 找出奇偶的第一和第二大
let [oddFirstNum, oddFirstCount, oddSecondNum, oddSecondCount] =
getValue(oddMap);
let [evenFirstNum, evenFirstCount, evenSecondNum, evenSecondCount] =
getValue(evenMap);
function getValue(map) {
let firstNum, firstCount = 0, secondNum, secondCount = 0;
for (let [item, value] of map) {
if (value >= firstCount) {
secondCount = firstCount;
firstCount = value;
secondNum = firstNum;
firstNum = item;
} else if (value < firstCount && value > secondCount) {
secondCount = value;
secondNum = item;
}
}
return [firstNum, firstCount, secondNum, secondCount]
}
// 最终要判断下奇偶最多的数是否相同
if (evenFirstNum !== oddFirstNum) {
return arr.length - oddFirstCount - evenFirstCount;
} else if (evenFirstNum === oddFirstNum) {
if (oddFirstCount - oddSecondCount >= evenFirstCount - evenSecondCount) {
return arr.length - oddFirstCount - evenSecondCount;
} else {
return arr.length - oddSecondCount - evenFirstCount;
}
}
}
console.log(getTimes([1, 1, 4, 5, 1, 4])); // 3 
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