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Solution.java
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264 lines (201 loc) · 7.46 KB
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package array;
import java.util.*;
public class Solution {
int Max = 0;
public int FindGreatestSumOfSubArray(int[] array) {
Fonction(array);
return Max;
}
public int Fonction(int[] array) {
if (array.length <= 0) return 0;
if (array.length == 1) return array[0];
int Fn_1_Max = Fonction(subArray(array)) + array[array.length - 1];
int arrayN = array[array.length - 1];
int max = Math.max(Fn_1_Max, arrayN);
if (max > Max) {
Max = max;
}
return max;
}
// 截取数组 末尾减一个
public int[] subArray(int[] origin) {
int[] newArray = new int[origin.length - 1];
for (int i = 0; i < newArray.length; i++) {
newArray[i] = origin[i];
}
return newArray;
}
// 找数组中出现次数超过一半的数
public int MoreThanHalfNum_Solution(int [] array) {
Map<Integer,Integer> ncMap = new HashMap<>();
for (int i = 0; i < array.length; i++) {
Integer countKey = ncMap.get(array[i]);
int count = countKey == null ? 1 :countKey+1;
ncMap.put(array[i],count);
if(count > array.length/2){
return array[i];
}
}
return 0;
}
//一个整型数组里除了两个数字之外，其他的数字都出现了两次。请写程序找出这两个只出现一次的数字。
public void FindNumsAppearOnce(int [] array,int num1[] , int num2[]) {
Map<Integer,Integer> ncMap = new HashMap<>();
for (int i = 0; i < array.length; i++) {
Integer countKey = ncMap.get(array[i]);
int count = countKey == null ? 1 :countKey+1;
ncMap.put(array[i],count);
}
ArrayList<Integer> list = new ArrayList<>();
for(Map.Entry<Integer,Integer> entry : ncMap.entrySet()){
if(entry.getValue() ==1){
list.add(entry.getKey());
}
}
num1[0] = list.get(0);
num2[0] = list.get(1);
}
class NumPo {
int num;
int position;
public NumPo(int num,int position){
this.num = num;
this.position = position;
}
@Override
public boolean equals(Object obj) {
NumPo numPo = (NumPo) obj;
return this.num == numPo.num && this.position != numPo.position;
}
}
//两数和问题
public ArrayList<Integer> FindNumbersWithSum(int [] array,int sum) {
ArrayList<Integer> result = new ArrayList<>();
if(array.length <2){
return result;
}
HashMap<Integer,ArrayList<Integer>> map = new HashMap<>();
for (int i = 0; i < array.length; i++) {
if(map.get(array[i])==null){
ArrayList<Integer> list = new ArrayList<>();
list.add(i);
map.put(array[i],list);
}else{
ArrayList<Integer> list = map.get(array[i]);
list.add(i);
map.put(array[i],list);
}
}
for (int i = 0; i < array.length; i++) {
int num2 = sum - array[i];
if(map.containsKey(num2)){
ArrayList<Integer> list = map.get(num2);
if(list.size() ==1 && list.get(0)== i){
continue;
}
result.add(array[i]);
result.add(num2);
break;
}
}
return result;
}
// 旋转数组中最小的那个
public int minNumberInRotateArray(int [] array) {
if(array.length == 0) return 0;
return RotateArray(array,0,array.length-1);
}
// // 旋转数组中最小的那个 构造递归循环
public int RotateArray(int[] array ,int low,int high){
int middle = (low+high)/2;
if (array[middle] < array[low]) return RotateArray(array,low,middle);
else if(array[middle] > array[low]) return RotateArray(array,middle,high);
else return array[high];
}
// 求次方
public double Power(double base, int exponent) {
if(exponent==0) return 1;
boolean sign = exponent >0 ? true: false;
double origin = base;
int ex = exponent >0 ? exponent: -exponent;
for (int i = 0; i < ex-1; i++) {
base*=origin;
System.out.println("====循环一次"+base);
}
return sign? base : 1/base;
}
/* public static void main(String[] args) {
int[] array = {4, -3, 5, -2, -1, 2, 6, -2};
Solution las = new Solution();
System.out.println(las.FindGreatestSumOfSubArray(array));
}*/
// 划分数组，奇前偶后
public void reOrderArray(int [] array) {
// int left_index = -1;
// // 遍历数组
// for (int i = 0; i < array.length ; i++) {
// if(array[i]%2 != 0){
// // 交换 left_index+1 和 i
// if(array[left_index+1]!= array[i]){
// array[left_index+1] = array[left_index+1]^ array[i];
// array[i] = array[left_index+1]^ array[i];
// array[left_index+1] = array[left_index+1]^ array[i];
// }
//
// left_index++;
// }
// }
int i= 0;
while (i<array.length){
while (i < array.length && array[i]%2 !=0) i++;
int first_ou = i;
while (i <array.length && array[i]%2 == 0) i++;
int first_ji = i;
if(first_ji == array.length || first_ou == array.length) continue;
// 第一个奇数和第一个偶数交换位置
if(array[first_ou]!= array[first_ji]){
System.out.println("first_ji :"+array[first_ji] +" first_ou:"+array[first_ou]);
array[first_ou] = array[first_ou]^ array[first_ji];
array[first_ji] = array[first_ou]^ array[first_ji];
array[first_ou] = array[first_ou]^ array[first_ji];
i++;
}
}
for (int j = 0; j < array.length; j++) {
System.out.print(array[j]+ " ");
}
}
/**
* 统计一个数字在排序数组中出现的次数
* @param array
* @param k
* @return
*/
public int GetNumberOfK(int [] array , int k) {
// 首先在数组中 二分查找k
int position = BinarySearch(array,k,0,array.length-1);
if(position == -1 ) return 0; // 没找到
int low = position;
int high = position;
// 通过高低指针移位并计数
while (low>=0 && array[low]== k) low--;
while (high < array.length && array[high] == k) high++;
return high-low-1;
}
public int BinarySearch(int [] array , int k ,int left,int right) {
if(left > right) return -1;
// if(left == right) return left;
int middle = (left+right) /2;
if(array[middle] == k ) return middle;
if(array[middle] > k) return BinarySearch(array,k,left,middle-1);
else return BinarySearch(array,k,middle+1,right);
}
public static void main(String[] args) {
int[] array = {3,3,3,3,4,5};
Solution solution = new Solution();
// System.out.println(solution.minNumberInRotateArray(array));
// solution.reOrderArray(array);
// System.out.println(solution.BinarySearch(array,3,0,array.length-1));
System.out.println(solution.GetNumberOfK(array,3));
}
}