how are you
Pre-knowledge
const Greedy = num => {
//贪心
let arr = [100, 20, 10, 5, 2, 1]
let count = 0;
for (let i = 0; i < arr.length; i++) {
let use = Math.floor(num / arr[i])
count += use;
num = num - arr[i] * use
console.log('需要面额为' + arr[i] + '有' + count + '张')
count = 0;
}
}
Greedy(120)A group of children eating candy
// 孩子,糖果,
const findContent = (g, s) => {
g.sort()
s.sort()
let child = 0;
let cookie = 0;
while (child < g.length && cookie < s.length) {
if (g[child] <= s[cookie]) {
child++
}
cookie++;//无论成功或者失败,每个糖果只尝试一次,cookie向后移动
}
return child
}
let g = [2, 4, 5, 6, 7]
let s = [1, 3, 4, 6, 7]
console.log(findContent(g, s))Swing sequence (LeetCode376)
for example
[1,17,5,10,13,15,10,5,16,8]
[5,10,13,15] belongs is incremented sequence
That is, [the small size of the big ...] or [Size Size ...]
Seeking the longest swing sequence
const wiggleMax = nums => { let n = nums.length if (n < 2) { return n } let up = 1; let down = 1; for (let i = 1; i <n ; i++) { if (nums[i] > nums[i - 1]) { up=down+1; } if (nums[i] < nums[i - 1]) { down=up+1; } } return Math.max(up, down) } console.log(wiggleMax([1, 17, 5, 10, 13, 15, 10, 5, 16, 8,1,9]))
402 k-bit digital definite displacement
Given a non-negative integer as a string of NUM , remove the number of k -bit numbers, the lowest number such that the remaining
Initializing a stack
Remember break point to understand
class Stack { constructor() { this.items = [] } push(element) { this.items.push(element) } //移除栈中的元素,遵循先进后出原则 pop() { return this.items.pop() } //返回栈顶 peek() { return this.items[this.items.length - 1] } //判断栈是否为空,为空为true,不为空为false isEmpty() { return this.items.length == 0 } size() { return this.items.length } //清除栈 clear() { this.items = []; } print() { console.log(this.items) } }
const removeK = (num, k) => {
//存在的栈
const stack = new Stack();
//贪心算法+栈
if (k >= num.length || num.length == 0) return '0';
//栈顶始终是最大值
stack.push(+num.charAt(0));
for (let i = 1; i < num.length; i++) {
let now = +num.charAt(i);
//当前栈不为空的时候,而且k>0,当前值小于栈顶,就删除栈中的一个元素
while (!stack.isEmpty() && k > 0 && now < stack.peek()) {
stack.pop();
k--;
}
//不等于0可以添加进去,
//等于0，栈不为空可以填进去，
if (now != 0 || !stack.isEmpty()) {
stack.push(now);
}
}
//56789去掉后面添加的个人
while (k > 0) {
k--;
stack.pop();
}
//10，1(当now=0时，满足条件，去掉1，但now为0，且为空。)
if (stack.isEmpty()) {
return '0';
}
//把栈中元素放到数组中
let sb = [];
while (!stack.isEmpty()) {
sb.push(stack.pop());
}
//还原成字符串
let sub = '';
sb.reverse().map(val => sub += val)
return sub
}
console.log(removeK('9234567', 3))Jumping game
Given a non-negative integer array, you initially positioned at the first position of the array.
Each element in the array represents the maximum length you can jump in that position.
Determine whether you can reach the last position of the array.
Sample
A = [2,3,1,1,4], returns true.A = [3,2,1,0,4], returns false.
const canJump = maxSteps => {
if (maxSteps == null || maxSteps.length == 0) {
return false;
}
let meetIndex = maxSteps.length - 1;
for (let i = maxSteps.length - 1; i >= 0; i--) {
if (i + maxSteps[i] >= meetIndex) {
meetIndex = i;
}
}
return meetIndex == 0;
}
console.log(canJump([0,2,0,1,4]))C ++ base
>> Assignment
<< print
-> .
Histogram largest rectangular area
const getAnswer = h => {
let n = h.length;
let ans = 0;
for (let i = 1; i < n; i++) {
let a = Infinity;
for (let j = i; j < n; j++) {
a = Math.min(a, h[j])
ans = Math.max(ans, (j - i + 1) * a)
}
}
return ans
}
console.log(getAnswer([1, 8, 3, 4, 8]))Please n items
Recursive version
const sum2=num=>{
if (num < 1) {
return 0
}
return sum2(num-1)+num
}
console.log(sum2(100))Non-recursive version
const sum1=n=>{
let result=0;
for (let i = 0; i <=n; i++) {
result+=i;
}
return result
}
console.log(sum1(100))Linear table
Is sequentially stored in the order table
Chain is a chain storage linear form, discontinuous
The address of the next element | data elements -
String
The memory may be used sequentially stored character string and chain stores two ways
BF algorithm: BF is brute force, brute force
BF algorithm is O (m * n)
const BF = (s, t, pos) => {
let i = pos,
j = 1,
sum = 0;
let slen = s.length
let tlen = s.length
while (i <= slen && j <= tlen) {
sum++
//如果相等,则继续比较后面的字符
if (s[i - 1] == t[j - 1]) {
i++
j++
} else {
//i回退到上一轮开始比较的下一个字符
i = i - j + 2
//j回退到第1个字符
j = 1;
}
}
return '一共比较了' + sum + '次'
}
console.log(BF('abcbcd', 'bcd', 0))KMP algorithm O (n + m)
Using dynamic programming to solve
True prefix except itself, all of a string head assembly
Prefix addition to its own, a string of all the tail combination
Animation
Calculating the maximum length of the common prefix and suffix of ( repeat length )
Start comparing (the array index to move forward 3 a)
Written mess, now under re-analysis
移动的位数=匹配的字符数-对应的部分匹配值// 4-3 = 1, a moveWe calculate the matching table exploded, p is represented by the prefix, n represents the suffix, r represents the results
a, p=>0, n=>0 r = 0 aa, p=>[a]，n=>[a] , r = a.length => 1 aar, p=>[a,aa], n=>[r,ar] ,r = 0 aaro, p=>[a,aa,aar], n=>[o,ra,aro] ,r = 0 aaron p=>[a,aa,aar,aaro], n=>[n,on,ron,aron] ,r = 0 aarona, p=>[a,aa,aar,aaro,aaron], n=>[a,na,ona,rona,arona] ,r = a.lenght = 1 aaronaa, p=>[a,aa,aar,aaro,aaron,aarona], n=>[a,aa,naa,onaa,ronaa,aronaa] , r = Math.max(a.length,aa.length) = 2 aaronaac p=>[a,aa,aar,aaro,aaron,aarona], n=>[c,ac,aac,naac,onaac,ronaac] r = 0Finally found a more comfortable looking at the complete code, not much to say directly on the code
const getNext = str => { let next = [-1] let k = -1 for (let i = 1; i < str.length; i++) { //第一次不执行且从前缀开始,判断不同的把-1赋值上 while (k != -1 && str[k + 1] != str[i]) { k = next[k] } //然后判断相同的,让k自增 if (str[k + 1] == str[i]) { k++ } //这是最开始将k赋值到数组中,然后依次判断把值赋值上 next[i] = k } return next } const KMP = (str1, str2) => { let next = getNext(str2) let j = 0 for (let i = 0; i < str1.length; i++) { while (j > 0 && str1[i] != str2[j]) { j = next[j - 1] + 1 // j 更新为最长可匹配前缀子串的长度 k } if (str1[i] == str2[j]) j++ if (j == str2.length) return i - str2.length + 1 } return -1 } console.log(KMP('abcdacbcdababc', 'ababc'))
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