leetcode 2 sum 3sum 4sum

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Two SumMar 14 '1115259 / 48327

Given an array of integers, find two numbers such that they add up to a specific target number.

The function twoSum should return indices of the two numbers such that they add up to the target, where index1 must be less than index2. Please note that your returned answers (both index1 and index2) are not zero-based.

You may assume that each input would have exactly one solution.

Input: numbers={2, 7, 11, 15}, target=9
Output: index1=1, index2=2


public class Solution {    public int[] twoSum(int[] numbers, int target) {        // Start typing your Java solution below        // DO NOT write main() function        if(numbers==null) return null;                int[] now = new int[numbers.length];        System.arraycopy(numbers,0,now,0,numbers.length);        Arrays.sort(numbers);        int i = 0,j=numbers.length-1;                while(i<j){            if(numbers[i]+numbers[j]>target){                j--;            }else if(numbers[i]+numbers[j]<target){                i++;            }else{                int x =0;                for(x=0;x<now.length;x++){                    if(now[x]==numbers[i])                    break;                }                //注意题目希望返回下标,所以要留意重新排序后造成的下标变化问题                int y=0;                for(y=0;y<now.length;y++){                    if(x!=y && now[y]==numbers[j])                    break;                }                /*                if(now[x]<now[y]){                    int temp = x;                    x=y;                    y=temp;                }*/                                int [] res = new int[2];                res[0]=x+1;                res[1]=y+1;                Arrays.sort(res);   //返回的下标是要有序的!                return res;            }                    }          return null;            }}
Run Status: Compile Error
Line 16: ']' expected
Run Status: Compile Error
Line 24: missing return statement
Run Status: Runtime Error
Last executed input
[5,75,25], 100
Run Status: 
Progress: 0/10 test cases passed.
Showing the first failed test case.

inputoutputexpected [5,75,25], 1001, 22, 3
 
Run Status: 
Progress: 2/10 test cases passed.
Showing the first failed test case.

inputoutputexpected [150,24,79,50,88,345,3], 2003, 61, 4
 
Run Status: Compile Error
Line 7: incompatible types
Run Status: Compile Error
Line 8: cannot find symbol: method copy(int[],int[])
Run Status: Compile Error
Line 8: cannot find symbol: method copyOf(int[],int[])
Run Status: Compile Error
Line 29: cannot find symbol: variable x
Run Status: 
Progress: 6/10 test cases passed.
Showing the first failed test case.

inputoutputexpected [5,75,25], 1003, 22, 3
 
Run Status: 
Progress: 6/10 test cases passed.
Showing the first failed test case.

inputoutputexpected [5,75,25], 1003, 22, 3
 
Run Status: 
Progress: 6/10 test cases passed.
Showing the first failed test case.

inputoutputexpected [5,75,25], 1003, 22, 3
 
Run Status: 
Progress: 3/10 test cases passed.
Showing the first failed test case.

inputoutputexpected [2,1,9,4,4,56,90,3], 84, 44, 5
 
Run Status: 
Progress: 4/10 test cases passed.
Showing the first failed test case.

inputoutputexpected [230,863,916,585,981,404,316,785,88,12,70,435,384,778,887,755,740,337,86,92,325,422,815,650,920,125,277,336,221,847,168,23,677,61,400,136,874,363,394,199,863,997,794,587,124,321,212,957,764,173,314,422,927,783,930,282,306,506,44,926,691,568,68,730,933,737,531,180,414,751,28,546,60,371,493,370,527,387,43,541,13,457,328,227,652,365,430,803,59,858,538,427,583,368,375,173,809,896,370,789], 54246, 2929, 46
 
Run Status: Accepted!
Program Runtime: 592 milli secs
Progress: 10/10 test cases passed.



3SumJan 18 '129830 / 38290

Given an array S of n integers, are there elements abc 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.

    For example, given array S = {-1 0 1 2 -1 -4},    A solution set is:    (-1, 0, 1)    (-1, -1, 2)
public class Solution {    public static ArrayList<ArrayList<Integer>> threeSum(int[] num) {// Start typing your Java solution below// DO NOT write main() functionif (num == null)return null;ArrayList<ArrayList<Integer>> list = new ArrayList<ArrayList<Integer>>();Arrays.sort(num);for (int i = 0; i < num.length-2; i++) {if (i > 0 && num[i] == num[i - 1])  continue;int j = i + 1, k = num.length - 1;while (j < k) {if (num[j] + num[k] > -num[i]) {k--;} else if (num[j] + num[k] < -num[i]) {j++; // 写错成i了....} else {ArrayList<Integer> a = new ArrayList<Integer>();a.add(num[i]);  //放的不是ijk 而是他们的值!a.add(num[j]); a.add(num[k]);list.add(a);                    j++;  //这些步骤是为了排除掉 j k里面存在重复的地方!                    k--;                       while(j<k && num[j]==num[j-1]) j++;                    while(j<k && num[k]==num[k+1])  k--;}}}return list;}}



Run Status: Compile Error
Line 27: illegal start of type
Run Status: Memory Limit Exceeded
Run Status: Memory Limit Exceeded
Run Status: Memory Limit Exceeded
Run Status: Memory Limit Exceeded
Run Status: 
Progress: 0/28 test cases passed.
Showing the first failed test case.

inputoutputexpected []null[]
 
Run Status: 
Progress: 12/28 test cases passed.
Showing the first failed test case.

inputoutputexpected [0,0,0][][[0,0,0]]
 
Run Status: 
Progress: 12/28 test cases passed.
Showing the first failed test case.

inputoutputexpected [0,0,0][[0,1,2]][[0,0,0]]
 
Run Status: Accepted!
Program Runtime: 532 milli secs
Progress: 28/28 test cases passed.


3Sum ClosestJan 18 '126070 / 15852

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.

    For example, given array S = {-1 2 1 -4}, and target = 1.    The sum that is closest to the target is 2. (-1 + 2 + 1 = 2).

public class Solution {    public int threeSumClosest(int[] num, int target) {        // Start typing your Java solution below        // DO NOT write main() function        if(num==null) return -1;                int goal = 0;  //不设为最大值! 因为加加减减会造成溢出!        boolean ifinit = false; //加入变量,来让goal有个初值!                Arrays.sort(num);        for(int i=0;i<num.length-2;i++){         if(i>0 && num[i]==num[i-1]) continue;         int j= i+1,k=num.length-1;                  while(j<k){             int now = num[i]+num[j]+num[k];             if(!ifinit||Math.abs(now-target)<Math.abs(goal-target) ){                  goal = now;                 ifinit = true;             }             if(now>target){                 k--;                    while(j<k && num[k]==num[k+1]) k--;  //前提条件!!!             }else{                 j++;                   while(j<k && num[j]==num[j-1]) j++;             }                    }        }        return goal;    }}

Run Status: Compile Error
Line 16: ')' expected
Run Status: Runtime Error
Last executed input
[0,0,0], 1
Run Status: 
Progress: 18/20 test cases passed.
Showing the first failed test case.

inputoutputexpected [1,1,-1,-1,3], -12147483647-1
 
Run Status: Compile Error
Line 17: cannot find symbol: variable init
Run Status: Compile Error
Line 17: variable goal might not have been initialized
Run Status: Accepted!
Program Runtime: 492 milli secs
Progress: 20/20 test cases passed.
Run Status: Accepted!
Program Runtime: 556 milli secs
Progress: 119/119 test cases passed.

  4sum  大数据超时  时间复杂度 n*3

 public static ArrayList<ArrayList<Integer>> fourSum(int[] num, int target) {        if(num==null) return null;                Arrays.sort(num);        ArrayList<ArrayList<Integer>>  list = new ArrayList<ArrayList<Integer>> ();                for(int i=0;i<num.length-3;i++){            if(i>0 && num[i]==num[i-1])  continue;                        for(int j=i+1;j<num.length-2;j++){                if(j>i+1 && num[j]==num[j-1]) continue;  //这里请注意j是从i+1开始的!!!                                int k = j+1,m = num.length-1;                while(k<m){                    if(num[i]+num[j]+num[k]+num[m]==target){                        ArrayList<Integer> a = new ArrayList<Integer>();                        a.add(num[i]);                        a.add(num[j]);                        a.add(num[k]);                        a.add(num[m]);                        list.add(a);                        k++;                        m--;                        while(k<m && num[k]==num[k-1]) k++;                        while(k<m && num[m]==num[m+1]) m--;                    }else if(num[i]+num[j]+num[k]+num[m]>target){                        m--;                    }else{                        k++;                    }                }            }        }        return list;    }