leetcode: 2Sum/3Sum/3SumClosest/4Sum系列问题

leetcode（http://leetcode.com/onlinejudge）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:

Elements in a triplet (a,b,c)
must be in non-descending order. (ie, a ? b ? c)

The solution set must not contain duplicate triplets.

有了twoSum的启发，threeSum所有做的事，只需加上排序，和一层循环。经测试AC。

import java.util.ArrayList;
import java.util.Arrays;

public class ThreeSumSolution2 {
private ArrayList<ArrayList<Integer>> list;

public ArrayList<ArrayList<Integer>> threeSum(int[] num) {
list = new ArrayList<ArrayList<Integer>>();
Arrays.sort(num);

int i = 0;
for (i = 0; i <= num.length - 3; i++) {
if (i != 0 && num[i] == num[i - 1]) {
continue;
}
judgeAndPut(num, i, i + 1, num.length - 1);
}

return list;
}

private void judgeAndPut(int[] num, int i, int p, int q) {
while (p < q) {
if (num[p] + num[q] < -num[i]) {
p++;
} else if (num[p] + num[q] > -num[i]){
q--;
} else if (num[p] + num[q] == -num[i]) {
ArrayList<Integer> tmpList = new ArrayList<Integer>();
tmpList.add(num[i]);
tmpList.add(num[p]);
tmpList.add(num[q]);
list.add(tmpList);
p++;
q--;
while (p < q && num[p] == num[p - 1]) {
p++;
}
while (p < q && num[q] == num[q + 1]) {
q--;
}
}
}
}

public static void main(String[] args) {
int num[] = {-1,0,1,2,-1,-4};
System.out.println(new ThreeSumSolution2().threeSum(num));
}
}

3. threeSumClosest: Given
an array S of n integers,
find three integers in S such
that the sum is closest to a given number, target. Return the sum of the three integers. You may assume that each input would have exactly one solution.
思路和threeSum几乎一样，只是查找条件有所变化而已。代码更简洁了。

import java.util.Arrays;

public class ThreeSumClosest {
private int closest;
private boolean needInit;

public int threeSumClosest(int[] num, int target) {
closest = 0;
needInit = true;
Arrays.sort(num);

int i = 0;
for (i = 0; i <= num.length - 3; i++) {
if (i != 0 && num[i] == num[i - 1]) {
continue;
}
judgeAndPut(num, i, i + 1, num.length - 1, target);
}
return closest;
}

private void judgeAndPut(int[] num, int i, int p, int q, int target) {

while (p < q) {
int sum = num[i] + num[p] + num[q];
if (needInit || Math.abs(sum - target) < Math.abs(closest - target)) {
closest = sum;
}
needInit = false;

if (sum <= target) {
p++;
while (p < q && num[p] == num[p - 1]) {
p++;
}
} else if (sum > target){
q--;
while (p < q && num[q] == num[q + 1]) {
q--;
}
}
}

}

public static void main(String[] args) {
int num[] = {0,1,2};
System.out.println(new ThreeSumClosest().threeSumClosest(num, 3));
}
}

4. fourSum:

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:

Elements in a quadruplet (a,b,c,d)
must be in non-descending order. (ie, a ? b ? c ? d)

The solution set must not contain duplicate quadruplets.

再多两层循环，此时已经是n立方的时间效率了，尝试了一下，居然也AC了。不知道还有没有更高效的算法。

import java.util.ArrayList;
import java.util.Arrays;

public class FourSum {
private ArrayList<ArrayList<Integer>> list;

public ArrayList<ArrayList<Integer>> fourSum(int[] num, int target) {
list = new ArrayList<ArrayList<Integer>>();
Arrays.sort(num);

int i = 0;
int j = 0;
for (i = 0; i <= num.length - 4; i++) {
if (i != 0 && num[i] == num[i - 1]) {
continue;
}

for (j = i + 1; j <= num.length - 3; j++) {
if (j != i + 1 && num[j] == num[j - 1]) {
continue;
}
judgeAndPut(num, i, j, j + 1, num.length - 1, target);
}

}

return list;
}

private void judgeAndPut(int[] num, int i, int j, int p, int q, int target) {
while (p < q) {
int sum = num[i] + num[j] + num[p] + num[q];
if (sum < target) {
p++;
} else if (sum > target){
q--;
} else if (sum == target) {
ArrayList<Integer> tmpList = new ArrayList<Integer>();
tmpList.add(num[i]);
tmpList.add(num[j]);
tmpList.add(num[p]);
tmpList.add(num[q]);
list.add(tmpList);
p++;
q--;
while (p < q && num[p] == num[p - 1]) {
p++;
}
while (p < q && num[q] == num[q + 1]) {
q--;
}
}
}

}

public static void main(String[] args) {
int num[] = {-1,0,1,0,-2,2};
System.out.println(new FourSum().fourSum(num, 0));

}

}