Open sisterAn opened 3 years ago
问题一
function queryFn(arr) {
if(!arr.length) return 0
let res = 1
let start = 0
for (let i = 1, len = arr.length; i < len; i++){
if (arr[i] < arr[i - 1]) {
res = Math.max(res,i-start)
start = i
}
}
return res
}
问题二
function queryFn(arr) {
if(!arr.length) return 0
let res = 1
let num = 0
let lastVal = 0
for (let i = 0, len = arr.length-1; i < len; i++){
for (let j = i + 1; j < len; j++){
if (arr[j] > lastVal) {
num++
lastVal = arr[j]
}
}
res = Math.max(res,num)
}
return res
}
const findLengthOfLCIS = (nums) => {
if(nums.length <= 1) return nums.length
let max = 1, count = 1
for(let i = 1; i < nums.length; i++){
if(nums[i] > nums[i-1]) {
count += 1
} else {
count = 1
}
max = Math.max(max, count)
}
return max
}
var maxLength = function(nums) {
if(nums.length <= 1) return nums.length
let count = 1, max = 1, start = 0
for(let i = 0; i < nums.length-1; i++) {
start = nums[i]
for(let j=i+1; j < nums.length; j++) {
if(nums[j]>start) {
count ++
start = nums[j]
}
}
max = Math.max(max, count)
count = 1
}
return max
}
var findLengthOfLCIS = function(nums) {
if (nums.length <= 1) return nums.length
let i = 0, j = 1
let max = 1
while (j < nums.length) {
if (nums[j] > nums[j - 1]) {
max = Math.max(max, j - i + 1)
j++
} else {
i = j
j = j + 1
}
}
return max
}
var findLengthOfLCIS = function(nums) {
if(nums.length <= 1) return nums.length
let max = 1,count = 1,len = nums.length
for(let i = 1; i < len; i++){
if(nums[i] > nums[i-1]){
count += 1
}else{
count = 1
}
max = Math.max(max,count)
}
return max
};
var findLengthOfLCIS = function(nums) {
if(nums.length <= 1) return nums.length
let max = 1,count = 1,len = nums.length
for(let i = 1; i < len; i++){
if(nums[i] !== (nums[i-1] + 1) && nums[i] > nums[i-1]){
count += 1
}else{
count = 1
}
max = Math.max(max,count)
}
return max
};
const 连续最大升序 = arr => {
let sum = 1, _sum = 1
for (let i = 0; i < arr.length - 1; i++) {
if (arr[i] < arr[i + 1]) _sum += 1
else _sum = 1
if (_sum > sum) sum = _sum
}
console.log(sum)
}
const 不连续最大升序 = arr => {
// 保存每个下标对应的最大升序数量
let 数组下标对应的非连续最大升序 = {}
// 从最后一个算
for (let i = arr.length - 1; i >= 0; i--) {
let 当前下标对应的所有非连续升序集 = [1]
for (let j = i + 1; j < arr.length - 1; j++) {
// 如果后面的值大于当前值 将累计值存入数据
if (arr[i] < arr[j]) 当前下标对应的所有非连续升序集.push(数组下标对应的非连续最大升序[j] + 1)
}
// 过滤出每个下标的最大值
数组下标对应的非连续最大升序[i] = Math.max(...当前下标对应的所有非连续升序集)
}
// 过滤出对象中最大的值
console.log(Math.max(...Object.values(数组下标对应的非连续最大升序)))
}
(arr)=>{
let index = 1
let temp = []
for(let i=0;i<arr.length;i++){
temp[i] = 1
for(let j = i ;j<arr.length;j++){
if(arr[j+1] > arr[j]){
temp[i] = ++temp[i]
if(temp[i]>index){
index++
}
}else{
break
}
}
}
}
let num = 1
let temp2 = []
for(let i=1;i<arr.length;i++){
temp2[i] = 1
for(let j = i ;j<arr.length;j++){
if(arr[j+1] > arr[j]){
temp2[i] = ++temp2[i]
if(temp2[i]>num){
num++
}
}else{
continue
}
}
}
解答问题一:
const findLengthOfLCIS = (nums) => { if(nums.length <= 1) return nums.length let max = 1, count = 1 for(let i = 1; i < nums.length; i++){ if(nums[i] > nums[i-1]) { count += 1 } else { count = 1 } max = Math.max(max, count) } return max }
解答问题二:@syc666
测试数据let arr = [101,19,12,34,51,32,7,103,8,104,1,105,2,106] 结果是6 实际结果应该是7吧 12 34 51 103 104 105 106
问题一
function queryFn(arr) { if(!arr.length) return 0 let res = 1 let start = 0 for (let i = 1, len = arr.length; i < len; i++){ if (arr[i] < arr[i - 1]) { res = Math.max(res,i-start) start = i } } return res }
问题二
function queryFn(arr) { if(!arr.length) return 0 let res = 1 let num = 0 let lastVal = 0 for (let i = 0, len = arr.length-1; i < len; i++){ for (let j = i + 1; j < len; j++){ if (arr[j] > lastVal) { num++ lastVal = arr[j] } } res = Math.max(res,num) } return res }
问题1 如果arr=[1,2,3]好像不正确呢
/**
* 连续不连续最大升序数量
* 连续可以使用双指针
* let arr = [101,19,12,34,51,32,7,103,8]
*/
//1. 连续升序最大数量
function findMaxAesc(arr) {
if (arr.length === 0) return 0
let res = 1
let slow = 0
for (let fast = 1, len = arr.length; fast < len; fast++) {
if (arr[fast] < arr[fast - 1]) {
slow = fast
}
res = Math.max(res, fast - slow + 1)
}
return res
}
function findMaxAesc3(arr) {
if (arr.length <= 1) return arr.length
let max = 1
let count = 1
for (let i = 1; i < arr.length; i++) {
if (arr[i] > arr[i - 1]) {
count++
} else {
count = 1
}
max = Math.max(max, count)
}
return max
}
// let arr = [101,19,12,34,51,32,7,103,8,104,1,105,2,106]
let arr = [2, 1, 3, 4, 5]
console.log(findMaxAesc(arr))
console.log(findMaxAesc3(arr))
//不连续最大升序数量
function findMaxAesc2(arr) {
if (arr.length === 0) return 0
let res = 1
for (let i = 0, leng = arr.length; i < leng; i++) {
let slow = arr[i]
let num = 1
for (let j = i + 1; j < leng; j++) {
if (arr[j] > slow) {
num++
slow = arr[j]
}
}
res = Math.max(res, num)
}
return res
}
console.log(findMaxAesc2(arr))
function fn(nums){ let maxcontinueUpCount=1,maxcontinueUpCountValue = 1; let maxBreakUpCount= 1,maxBreakUpCountValue =1 let i = 1 while(i<nums.length){ if(nums[i]>nums[i-1]){ if(maxBreakUpCountValue<maxBreakUpCount){ maxBreakUpCountValue = maxBreakUpCount } maxBreakUpCount=1 maxcontinueUpCount++ }else{ if(maxcontinueUpCountValue<maxcontinueUpCount){ maxcontinueUpCountValue = maxcontinueUpCount } maxcontinueUpCount = 1 maxBreakUpCount++ } i++ } return { maxBreakUpCountValue, maxcontinueUpCountValue } }
console.log(fn([101,19,12,51,32,7,103,8]))
问题一: 找出连续最大升序的数量
问题二: 找出不连续最大升序的数量