标签：des c style class blog code
转载：http://blog.csdn.net/li4951/article/details/8693212
leetcode上有好几道关于数组中几个数据和为target的题目。恰好正在看剑指offer中“和为s的两个数组这章”，据此思想，leetcode上的三道题目都被我解决了。总结一下。
1.twoSum:
输入一个递增数组和一个数字s，在数组中查找两个数使得它们的和正好是s。
既然题目中已经提到了“递增数组”,那么肯定不会暴力了。因此肯定有<O(n*n)的办法了。
算法：最初我们找到数组的第一个数字和最后一个数字。当两个数字的和大于输入的数字时，把较大的数字往前移动；当两个数字的和小于数字时，把较小的数字往后移动；当相等时，打完收工。这样扫描的顺序是从数组的两端向数组的中间扫描。
链接：http://zhedahht.blog.163.com/blog/static/2541117420072143251809/
2.threeSum:
Given an array S of n integers, are there elements a, b, c in S such that a + b + c = 0? Find all unique triplets in the array which gives the sum of zero.
Note:
1 import java.util.ArrayList; 2 import java.util.Arrays; 3 4 public class ThreeSumSolution2 { 5 private ArrayList<ArrayList<Integer>> list; 6 7 public ArrayList<ArrayList<Integer>> threeSum(int[] num) { 8 list = new ArrayList<ArrayList<Integer>>(); 9 Arrays.sort(num); 10 11 int i = 0; 12 for (i = 0; i <= num.length - 3; i++) { 13 if (i != 0 && num[i] == num[i - 1]) { 14 continue; 15 } 16 judgeAndPut(num, i, i + 1, num.length - 1); 17 } 18 19 return list; 20 } 21 22 private void judgeAndPut(int[] num, int i, int p, int q) { 23 while (p < q) { 24 if (num[p] + num[q] < -num[i]) { 25 p++; 26 } else if (num[p] + num[q] > -num[i]){ 27 q--; 28 } else if (num[p] + num[q] == -num[i]) { 29 ArrayList<Integer> tmpList = new ArrayList<Integer>(); 30 tmpList.add(num[i]); 31 tmpList.add(num[p]); 32 tmpList.add(num[q]); 33 list.add(tmpList); 34 p++; 35 q--; 36 while (p < q && num[p] == num[p - 1]) { 37 p++; 38 } 39 while (p < q && num[q] == num[q + 1]) { 40 q--; 41 } 42 } 43 } 44 } 45 46 public static void main(String[] args) { 47 int num[] = {-1,0,1,2,-1,-4}; 48 System.out.println(new ThreeSumSolution2().threeSum(num)); 49 } 50 }
1 import java.util.Arrays; 2 3 public class ThreeSumClosest { 4 private int closest; 5 private boolean needInit; 6 7 public int threeSumClosest(int[] num, int target) { 8 closest = 0; 9 needInit = true; 10 Arrays.sort(num); 11 12 int i = 0; 13 for (i = 0; i <= num.length - 3; i++) { 14 if (i != 0 && num[i] == num[i - 1]) { 15 continue; 16 } 17 judgeAndPut(num, i, i + 1, num.length - 1, target); 18 } 19 return closest; 20 } 21 22 private void judgeAndPut(int[] num, int i, int p, int q, int target) { 23 24 while (p < q) { 25 int sum = num[i] + num[p] + num[q]; 26 if (needInit || Math.abs(sum - target) < Math.abs(closest - target)) { 27 closest = sum; 28 } 29 needInit = false; 30 31 if (sum <= target) { 32 p++; 33 while (p < q && num[p] == num[p - 1]) { 34 p++; 35 } 36 } else if (sum > target){ 37 q--; 38 while (p < q && num[q] == num[q + 1]) { 39 q--; 40 } 41 } 42 } 43 44 } 45 46 public static void main(String[] args) { 47 int num[] = {0,1,2}; 48 System.out.println(new ThreeSumClosest().threeSumClosest(num, 3)); 49 } 50 }
Given an array S of n integers, are there elements a, b, c, and d in S such that a + b + c + d = target? Find all unique quadruplets in the array which gives the sum of target.
Note:
1 import java.util.ArrayList; 2 import java.util.Arrays; 3 4 public class FourSum { 5 private ArrayList<ArrayList<Integer>> list; 6 7 public ArrayList<ArrayList<Integer>> fourSum(int[] num, int target) { 8 list = new ArrayList<ArrayList<Integer>>(); 9 Arrays.sort(num); 10 11 int i = 0; 12 int j = 0; 13 for (i = 0; i <= num.length - 4; i++) { 14 if (i != 0 && num[i] == num[i - 1]) { 15 continue; 16 } 17 18 for (j = i + 1; j <= num.length - 3; j++) { 19 if (j != i + 1 && num[j] == num[j - 1]) { 20 continue; 21 } 22 judgeAndPut(num, i, j, j + 1, num.length - 1, target); 23 } 24 25 } 26 27 return list; 28 } 29 30 private void judgeAndPut(int[] num, int i, int j, int p, int q, int target) { 31 while (p < q) { 32 int sum = num[i] + num[j] + num[p] + num[q]; 33 if (sum < target) { 34 p++; 35 } else if (sum > target){ 36 q--; 37 } else if (sum == target) { 38 ArrayList<Integer> tmpList = new ArrayList<Integer>(); 39 tmpList.add(num[i]); 40 tmpList.add(num[j]); 41 tmpList.add(num[p]); 42 tmpList.add(num[q]); 43 list.add(tmpList); 44 p++; 45 q--; 46 while (p < q && num[p] == num[p - 1]) { 47 p++; 48 } 49 while (p < q && num[q] == num[q + 1]) { 50 q--; 51 } 52 } 53 } 54 55 } 56 57 public static void main(String[] args) { 58 int num[] = {-1,0,1,0,-2,2}; 59 System.out.println(new FourSum().fourSum(num, 0)); 60 61 } 62 63 }
leetcode: 2Sum/3Sum/3SumClosest/4Sum系列问题(转载),布布扣,bubuko.com
leetcode: 2Sum/3Sum/3SumClosest/4Sum系列问题(转载)
标签：des c style class blog code
原文地址：http://www.cnblogs.com/wolohaha/p/3778591.html
