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Oct 18, 2018
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018.四数之和[Java] #25

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018.四数之和[Java]
Mrzhudky Oct 17, 2018
dbac5be
Add solution 018[Java]
Mrzhudky Oct 18, 2018
102b19e
Update format and solution description For Solution 018[Java]
Mrzhudky Oct 18, 2018
408e8e9
Repair list display problem
Mrzhudky Oct 18, 2018
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Add solution 018[Java]
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@Mrzhudky
Mrzhudky committed Oct 18, 2018
commit dbac5bebdc4fe5ee19a3d77b7e2a32b1a8e9596d
35 changes: 19 additions & 16 deletions solution/018.4Sum/README.md
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Expand Up @@ -23,70 +23,73 @@


### 解法
先将4Sum问题，转换为3Sum问题，再转换为2Sum问题.
1、将数组排序
2、先假设确定一个数 nums[i] 将 4Sum 问题转换为 3Sum 问题；
3、再假设确定一个数将 3Sum 问题转换为 2Sum 问题；
4、对排序数组，用首尾指针向中间靠拢的思路寻找满足 target 的 nums[l] 和 nums[k]

```java
class Solution {
public List<List<Integer>> fourSum(int[] nums, int target) {

List<List<Integer>> re = new ArrayList<>();
if(nums == null || nums.length<4){
if(nums == null || nums.length<4) {
return re;
}
Arrays.sort(nums);
for (int i = 0; i < nums.length-3; i++) {

//当 nums[i] 对应的最小组合都大于 target 时，后面大于 nums[i] 的组合必然也大于 target，
if(nums[i]+nums[i+1]+nums[i+2]+nums[i+3] > target){
// 当 nums[i] 对应的最小组合都大于 target 时，后面大于 nums[i] 的组合必然也大于 target，
if(nums[i]+nums[i+1]+nums[i+2]+nums[i+3] > target) {
break;
}
// 当 nums[i] 对应的最大组合都小于 target 时， nums[i] 的其他组合必然也小于 target
if(nums[i]+nums[nums.length-3]+nums[nums.length-2] + nums[nums.length-1] < target){
if(nums[i]+nums[nums.length-3]+nums[nums.length-2] + nums[nums.length-1] < target) {
continue;
}

int firstNum = nums[i];
for (int j = i+1; j < nums.length-2; j++) {

// nums[j] 过大时，与nums[i]过大同理
if(nums[i]+nums[j]+nums[j+1]+nums[j+2] > target){
// nums[j] 过大时，与 nums[i] 过大同理
if(nums[i] + nums[j] + nums[j+1] + nums[j+2] > target) {
break;
}
// nums[j] 过小时，与nums[i]过小同理
if(nums[i]+nums[j]+nums[nums.length-2] + nums[nums.length-1] < target){
// nums[j] 过小时，与 nums[i] 过小同理
if(nums[i] + nums[j] + nums[nums.length-2] + nums[nums.length-1] < target) {
continue;
}

int twoSum = target - nums[i]-nums[j];
int l = j+1;
int k = nums.length-1;
while (l<k){
while (l<k) {
int tempSum = nums[l] + nums[k];
if(tempSum == twoSum){
if(tempSum == twoSum) {
ArrayList<Integer> oneGroup = new ArrayList<>(4);
oneGroup.add(nums[i]);
oneGroup.add(nums[j]);
oneGroup.add(nums[l++]);
oneGroup.add(nums[k--]);
re.add(oneGroup);
while (l<nums.length && l<k && nums[l] == oneGroup.get(2) && nums[k] == oneGroup.get(3)){
while (l < nums.length && l < k && nums[l] == oneGroup.get(2) && nums[k] == oneGroup.get(3)) {
l++;k--;
}
}
else if(tempSum<twoSum){
else if(tempSum < twoSum) {
l++;
}
else{
else {
k--;
}
}
// 跳过重复项
while ((j<nums.length-2)&&(twoSum + nums[i] + nums[j+1] == target)){
while ((j < nums.length - 2) && (twoSum + nums[i] + nums[j+1] == target)) {
j++;
}
}
// 跳过重复项
while (i<nums.length-3 &&(nums[i+1] == firstNum) ){
while (i < nums.length-3 && nums[i+1] == firstNum){
i++;
}
}
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30 changes: 15 additions & 15 deletions solution/018.4Sum/Solution.java
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Original file line number Diff line number Diff line change
Expand Up @@ -2,63 +2,63 @@ class Solution {
public List<List<Integer>> fourSum(int[] nums, int target) {

List<List<Integer>> re = new ArrayList<>();
if(nums == null || nums.length<4){
if(nums == null || nums.length<4) {
return re;
}
Arrays.sort(nums);
for (int i = 0; i < nums.length-3; i++) {

//当 nums[i] 对应的最小组合都大于 target 时，后面大于 nums[i] 的组合必然也大于 target，
if(nums[i]+nums[i+1]+nums[i+2]+nums[i+3] > target){
// 当 nums[i] 对应的最小组合都大于 target 时，后面大于 nums[i] 的组合必然也大于 target，
if(nums[i]+nums[i+1]+nums[i+2]+nums[i+3] > target) {
break;
}
// 当 nums[i] 对应的最大组合都小于 target 时， nums[i] 的其他组合必然也小于 target
if(nums[i]+nums[nums.length-3]+nums[nums.length-2] + nums[nums.length-1] < target){
if(nums[i]+nums[nums.length-3]+nums[nums.length-2] + nums[nums.length-1] < target) {
continue;
}

int firstNum = nums[i];
for (int j = i+1; j < nums.length-2; j++) {

// nums[j] 过大时，与nums[i]过大同理
if(nums[i]+nums[j]+nums[j+1]+nums[j+2] > target){
// nums[j] 过大时，与 nums[i] 过大同理
if(nums[i] + nums[j] + nums[j+1] + nums[j+2] > target) {
break;
}
// nums[j] 过小时，与nums[i]过小同理
if(nums[i]+nums[j]+nums[nums.length-2] + nums[nums.length-1] < target){
// nums[j] 过小时，与 nums[i] 过小同理
if(nums[i] + nums[j] + nums[nums.length-2] + nums[nums.length-1] < target) {
continue;
}

int twoSum = target - nums[i]-nums[j];
int l = j+1;
int k = nums.length-1;
while (l<k){
while (l<k) {
int tempSum = nums[l] + nums[k];
if(tempSum == twoSum){
if(tempSum == twoSum) {
ArrayList<Integer> oneGroup = new ArrayList<>(4);
oneGroup.add(nums[i]);
oneGroup.add(nums[j]);
oneGroup.add(nums[l++]);
oneGroup.add(nums[k--]);
re.add(oneGroup);
while (l<nums.length && l<k && nums[l] == oneGroup.get(2) && nums[k] == oneGroup.get(3)){
while (l < nums.length && l < k && nums[l] == oneGroup.get(2) && nums[k] == oneGroup.get(3)) {
l++;k--;
}
}
else if(tempSum<twoSum){
else if(tempSum < twoSum) {
l++;
}
else{
else {
k--;
}
}
// 跳过重复项
while ((j<nums.length-2)&&(twoSum + nums[i] + nums[j+1] == target)){
while ((j < nums.length - 2) && (twoSum + nums[i] + nums[j+1] == target)) {
j++;
}
}
// 跳过重复项
while (i<nums.length-3 &&(nums[i+1] == firstNum) ){
while (i < nums.length-3 && nums[i+1] == firstNum){
i++;
}
}
Expand Down