8.位操作符：
从一个给定的数n中找位i（i从0开始，然后向右开始）
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public static boolean getBit(int num, int i){ int result = num & (1<<i); if(result == 0){ return false; }else{ return true; }} |
例如，获取10的第二位：
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i=1, n=101<<1= 101010&10=1010 is not 0, so return true; |
典型的位算法：
Find Single Number
The problem:
Given an array of integers, every element appears twice except for one. Find that single one.
Thoughts
The key to solve this problem is bit manipulation. XOR will return 1 only on two different bits. So if two numbers are the same, XOR will return 0. Finally only one number left.
Java Solution
public class Solution { public int singleNumber(int[] A) { int x=0; for(int a: A){ x = x ^ a; } return x; } }
Maximum Binary Gap
Problem: Get maximum binary Gap.
For example, 9′s binary form is 1001, the gap is 2.
Thoughts
The key to solve this problem is the fact that an integer x & 1 will get the last digit of the integer.
Java Solution
public class Solution {
public static int solution(int N) {
int max = 0;
int count = -1;
int r = 0;
while (N > 0) {
// get right most bit & shift right
r = N & 1;
N = N >> 1;
if (0 == r && count >= 0) {
count++;
}
if (1 == r) {
max = count > max ? count : max;
count = 0;
}
}
return max;
}
public static void main(String[] args) {
System.out.println(solution(9));
}
}
9.概率
通常要解决概率相关问题，都需要很好地格式化问题，下面提供一个简单的例子：
有50个人在一个房间，那么有两个人是同一天生日的可能性有多大？（忽略闰年，即一年有365天）
算法：
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public static double caculateProbability(int n){ double x = 1; for(int i=0; i<n; i++){ x *= (365.0-i)/365.0; } double pro = Math.round((1-x) * 100); return pro/100;} |
结果：
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calculateProbability(50) = 0.97 |
10.组合和排列
组合和排列的主要差别在于顺序是否重要。
例1：
1、2、3、4、5这5个数字，输出不同的顺序，其中4不可以排在第三位，3和5不能相邻，请问有多少种组合？
例2：
有5个香蕉、4个梨、3个苹果，假设每种水果都是一样的，请问有多少种不同的组合？
基于它们的一些常见算法
排列1
Given a collection of numbers, return all possible permutations.
For example,
[1,2,3] have the following permutations:
[1,2,3], [1,3,2], [2,1,3], [2,3,1], [3,1,2], and [3,2,1].
Java Solution 1
We can get all permutations by the following steps:
[1]
[2, 1]
[1, 2]
[3, 2, 1]
[2, 3, 1]
[2, 1, 3]
[3, 1, 2]
[1, 3, 2]
[1, 2, 3]
Loop through the array, in each iteration, a new number is added to different locations of results of previous iteration. Start from an empty List.
public ArrayList<ArrayList<Integer>> permute(int[] num) {
ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();
//start from an empty list
result.add(new ArrayList<Integer>());
for (int i = 0; i < num.length; i++) {
//list of list in current iteration of the array num
ArrayList<ArrayList<Integer>> current = new ArrayList<ArrayList<Integer>>();
for (ArrayList<Integer> l : result) {
// # of locations to insert is largest index + 1
for (int j = 0; j < l.size()+1; j++) {
// + add num[i] to different locations
l.add(j, num[i]);
ArrayList<Integer> temp = new ArrayList<Integer>(l);
current.add(temp);
//System.out.println(temp);
// - remove num[i] add
l.remove(j);
}
}
result = new ArrayList<ArrayList<Integer>>(current);
}
return result;
}
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Java Solution 2
We can also recursively solve this problem. Swap each element with each element after it.
public ArrayList<ArrayList<Integer>> permute(int[] num) {
ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();
permute(num, 0, result);
return result;
}
void permute(int[] num, int start, ArrayList<ArrayList<Integer>> result) {
if (start >= num.length) {
ArrayList<Integer> item = convertArrayToList(num);
result.add(item);
}
for (int j = start; j <= num.length - 1; j++) {
swap(num, start, j);
permute(num, start + 1, result);
swap(num, start, j);
}
}
private ArrayList<Integer> convertArrayToList(int[] num) {
ArrayList<Integer> item = new ArrayList<Integer>();
for (int h = 0; h < num.length; h++) {
item.add(num[h]);
}
return item;
}
private void swap(int[] a, int i, int j) {
int temp = a[i];
a[i] = a[j];
a[j] = temp;
}
排列2
Given a collection of numbers that might contain duplicates, return all possible unique permutations.
For example,
[1,1,2] have the following unique permutations:
[1,1,2], [1,2,1], and [2,1,1].
Thoughts
Basic idea: For each number in the array, swap it with every element after it. To avoid duplicate, need to check it first.
Java Solution
public ArrayList<ArrayList<Integer>> permuteUnique(int[] num) { ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>(); permuteUnique(num, 0, result); return result; } private void permuteUnique(int[] num, int start, ArrayList<ArrayList<Integer>> result) { if (start >= num.length ) { ArrayList<Integer> item = convertArrayToList(num); result.add(item); } for (int j = start; j <= num.length-1; j++) { if (containsDuplicate(num, start, j)) { swap(num, start, j); permuteUnique(num, start + 1, result); swap(num, start, j); } } } private ArrayList<Integer> convertArrayToList(int[] num) { ArrayList<Integer> item = new ArrayList<Integer>(); for (int h = 0; h < num.length; h++) { item.add(num[h]); } return item; } private boolean containsDuplicate(int[] arr, int start, int end) { for (int i = start; i <= end-1; i++) { if (arr[i] == arr[end]) { return false; } } return true; } private void swap(int[] a, int i, int j) { int temp = a[i]; a[i] = a[j]; a[j] = temp; }
排列顺序
The set [1,2,3,…,n] contains a total of n! unique permutations.
By listing and labeling all of the permutations in order,
We get the following sequence (ie, for n = 3):
"123"
"132"
"213"
"231"
"312"
"321"
Given n and k, return the kth permutation sequence.
Note: Given n will be between 1 and 9 inclusive.
Thoughts
Naively loop through all cases will not work.
Java Solution 1
public class Solution {
public String getPermutation(int n, int k) {
// initialize all numbers
ArrayList<Integer> numberList = new ArrayList<Integer>();
for (int i = 1; i <= n; i++) {
numberList.add(i);
}
// change k to be index
k--;
// set factorial of n
int mod = 1;
for (int i = 1; i <= n; i++) {
mod = mod * i;
}
String result = "";
// find sequence
for (int i = 0; i < n; i++) {
mod = mod / (n - i);
// find the right number(curIndex) of
int curIndex = k / mod;
// update k
k = k % mod;
// get number according to curIndex
result += numberList.get(curIndex);
// remove from list
numberList.remove(curIndex);
}
return result.toString();
}
}
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Java Solution 2
public class Solution {
public String getPermutation(int n, int k) {
boolean[] output = new boolean[n];
StringBuilder buf = new StringBuilder("");
int[] res = new int[n];
res[0] = 1;
for (int i = 1; i < n; i++)
res[i] = res[i - 1] * i;
for (int i = n - 1; i >= 0; i--) {
int s = 1;
while (k > res[i]) {
s++;
k = k - res[i];
}
for (int j = 0; j < n; j++) {
if (j + 1 <= s && output[j]) {
s++;
}
}
output[s - 1] = true;
buf.append(Integer.toString(s));
}
return buf.toString();
}
}
