编程之美2.12 快速寻找满足条件的两个数

//题目：给一个数组，求数组中两个数字之和等于给定数字的元素  //解法1：穷举法，计算两两元素的和是否满足要求，时间复杂度O(n^2)  public class Main {   public static void main(String[] args) {       findNum(new int[]{5,6,8,3,7,9,4},12);   }   public static void findNum(int[] input, int target) {       for(int i = 0;i<input.length;i++){           for(int j = i+1;j<input.length;j++){               if(input[i]+input[j] == target){                   System.out.println(input[i]+" "+input[j]);               }           }       }   }     }  //解法2：先排序，再使数组第一个元素与最后一个元素求和，根据sum与target关系调整low和high的位置，时间复杂度O(NlogN)因为要排序  public class Main {   public static void main(String[] args) {       findNum(new int[]{5,6,8,3,7,9,4},12);   }   public static void findNum(int[] input, int target) {       sort(input,0,input.length-1);       int low = 0;       int high = input.length-1;       while(low<high){           int sum = input[low]+input[high];           if(sum == target){               System.out.println(input[low]+" "+input[high]);               high--;           }else if(sum>target){               high--;           }else{               low++;           }       }   }      public static void sort(int[] input, int low, int high){       if(low<high){           int mid = partition(input, low, high);           sort(input, low, mid-1);           sort(input, mid+1, high);       }   }      public static int partition(int[] input, int low, int high){       int temp = input[low];       while(low<high){           while(low<high && input[high]>temp){               high--;           }           input[low] = input[high];           while(low<high && input[low]<temp){               low++;           }           input[high] = input[low];       }       input[low] = temp;       return low;   }  }  //解法3：hash表，以空间换时间，时间复杂度O(N),空间复杂度O(N)  public class Main {   public static void main(String[] args) {       findNum(new int[]{5,6,8,3,7,9,4},12);   }   public static void findNum(int[] input, int target) {       HashSet<Integer> hash = new HashSet<Integer>();       HashSet<Integer> hash1 = new HashSet<Integer>();    //避免重复，3+9 = 12，9+3 = 12       for(int i = 0;i<input.length;i++){           if(!hash.contains(input[i])){               hash.add(input[i]);           }       }       for(int i = 0;i<input.length;i++){           int value = target-input[i];           hash.remove(input[i]);                  //防止6+6 = 12这种情况发生           if(!hash1.contains(value) && hash.contains(value)){               System.out.println(input[i]+" "+value);           }           hash.add(input[i]);           hash1.add(input[i]);       }   }     }

// 扩展题目：给一个数组，求其中n个数字和为指定数字的n个元素// 解法：按照循环的方式找前n-2个元素，最后两个元素用之前的方法找public class Main {public static void main(String[] args) {List<List<Integer>> result = new ArrayList<List<Integer>>();List<Integer> temp = new ArrayList<Integer>();findNum(new int[]{1,2,3,4,5,6,7,8,9},20,4,temp,result);for(int i = 0;i<result.size();i++){printList(result.get(i));}}public static void findNum(int[] input, int target, int n, List<Integer> temp,List<List<Integer>> result) {if(n == 2){int low = 0;int high = input.length-1;while(low<high){if(target == input[low]+input[high]){temp.add(input[low]);temp.add(input[high]);if(!isContain(result,temp)){result.add(new ArrayList<Integer>(temp));}return;}else if(target > input[low]+input[high]){low++;}else{high--;}}return;}else{for(int i = 0;i<input.length;i++){temp.add(input[i]);int[] input_temp = copyarry(input,i);findNum(input_temp,target-input[i],n-1,temp,result);if(temp.size() == 4){temp.remove(temp.size()-1);temp.remove(temp.size()-1);temp.remove(temp.size()-1);}else{temp.remove(temp.size()-1);}}}}private static int[] copyarry(int[] arry, int i) {  //在数组中删除出现过的元素        int[] tmp = new int[arry.length - 1];          int idx = 0;          for (int j = 0; j < arry.length; j++) {                if (j != i) {                  tmp[idx++] = arry[j];              }          }          return tmp;      }  public static void printList(List<Integer> list){//打印列表中元素for(int i = 0;i<list.size();i++){System.out.print(list.get(i)+" ");}System.out.println();}public static boolean isContain(List<List<Integer>> result, List<Integer> temp){//判断result中是否有与temp一样的组合出现for(int i = 0;i<result.size();i++){List<Integer> t = new ArrayList<Integer>();for(int j = 0;j<result.get(i).size();j++){t.add(result.get(i).get(j));}for(int j = 0;j<temp.size();j++){if(t.contains(temp.get(j))){t.remove(temp.get(j));}}if(t.size() == 0){return true;}}return false;}}