*LeetCode-4Sum
o（n^3）
固定一个数字 然后就变成3sum 再固定一个就是2sum
但是注意处理duplicate 就是四个数字都要避免dup 处理了四个地方
public class Solution {
public List<List<Integer>> fourSum(int[] nums, int target) {
List<List<Integer>> res = new ArrayList<List<Integer>> ();
if ( nums == null )
return res;
Arrays.sort ( nums );
for ( int i = 0; i <= nums.length - 4; i ++ ){
int target3 = target - nums [ i ];
for ( int j = i + 1; j <= nums.length - 3; j ++ ){
int target2 = target3 - nums[ j ];
int head = j + 1;
int tail = nums.length - 1;
while ( head < tail ){
if ( nums [ head ] + nums[ tail ] < target2 )
head ++;
else if ( nums [ head ] + nums [ tail ] > target2 )
tail --;
else{
List<Integer> four = new ArrayList<Integer> ();
four.add ( nums [ i ] );
four.add ( nums [ j ] );
four.add ( nums [ head ] );
four.add ( nums [ tail ] );
res.add ( four );
while ( head < tail && nums[ head ] == four.get ( 2 ) )
head ++;
while ( head < tail && nums [ tail ] == four.get ( 3 ) )
tail --;
}
}
while ( j + 1 <= nums.length - 3 && nums [ j + 1 ] == nums [ j ] ) j ++;
}
while ( i + 1 <= nums.length - 4 && nums [ i + 1 ] == nums [ i ] ) i ++;
}
return res;
}
}