[leetCode]169.多数元素
暴力法
两层循环通过遍历判断每个数字的出现次数是否大于floor
class Solution {
public int majorityElement(int[] nums) {
int floor = nums.length / 2;
for(int i = 0; i < nums.length; i++){
int count = 0;
for(int j = 0; j < nums.length; j++){
if(nums[i] == nums[j]){
++count;
}
}
if(count > floor) return nums[i];
}
return 0;
}
}
哈希表
用哈希表记录对应元素与其出现的次数
class Solution {
public int majorityElement(int[] nums) {
int floor = nums.length/2;
HashMap<Integer, Integer> map = new HashMap<>();
for(int i = 0; i < nums.length; i++){
if(!map.keySet().contains(nums[i])){
map.put(nums[i],1);
}else {
map.replace(nums[i], map.get(nums[i])+1);
}
}
for(int i = 0; i < nums.length; i++){
if(map.get(nums[i])>floor) return nums[i];
}
return 0;
}
}
```java
class Solution {
public int majorityElement(int[] nums) {
HashMap<Integer, Integer> map = new HashMap<>();
int ans = nums[0];
int count = 1;
for(int i = 0; i < nums.length; i++){
if(!map.keySet().contains(nums[i])){
map.put(nums[i],1);
}else {
map.replace(nums[i], map.get(nums[i])+1);
if(map.get(nums[i])+1 > count){
count = map.get(nums[i])+1;
ans = nums[i];
}
}
}
return ans;
}
}
双指针
先排序,再利用双指针计数
class Solution {
public int majorityElement(int[] nums) {
if(nums.length == 1)return nums[0];
int floor = nums.length/2;
Arrays.sort(nums);
int ans = 0;
int count = 0;
int i = 0;
for(int j = i + 1; j < nums.length; j++){
if(nums[i] != nums[j] && j-i>count) {
count = j-i;
ans = nums[i];
i = j;
}
if(j == nums.length-1 && j-i+1 > count){
ans = nums[i];
}
}
return ans;
}
}
排序
众数肯定大于数组的一半,所以对数组排序,返回中间值即可
class Solution {
public int majorityElement(int[] nums) {
Arrays.sort(nums);
return nums[nums.length/2];
}
}
随机化
众数占多数,通过随机选取一个数字,判断该数是否为众数。如果是则退出
class Solution {
public int randIndex(Random rand, int min, int max){
return rand.nextInt(max - min) + min;
}
public int countOccurence(int x, int[] nums){
int count = 0;
for(int i = 0; i < nums.length; i++){
if(nums[i] == x) ++count;
}
return count;
}
public int majorityElement(int[] nums) {
Random rand = new Random();
while(true){
int current = nums[randIndex(rand, 0, nums.length)];
if(countOccurence(current, nums) > nums.length /2){
return current;
}
}
}
}
分治
如果数 a 是数组 nums 的众数,如果我们将 nums 分成两部分,那么 a 必定是至少一部分的众数。
class Solution {
public int countInRange(int x,int[] nums ,int lo, int hi){
int count = 0;
for(int i = lo; i <= hi; i++){
if(nums[i] == x) ++count;
}
return count;
}
public int majorityElementRec(int[] nums, int lo, int hi){
if(lo == hi) return nums[lo];
int mid = lo + (hi - lo)/2;
int left = majorityElementRec(nums, lo, mid);
int right = majorityElementRec(nums,mid+1,hi);
if(left == right) return left;
if(countInRange(left,nums,lo,hi) > countInRange(right,nums,lo,hi)){
return left;
}else{
return right;
}
}
public int majorityElement(int[] nums) {
return majorityElementRec(nums, 0, nums.length -1);
}
}
Boyer-Moore 投票算法
class Solution {
public int majorityElement(int[] nums) {
int count = 0;
Integer candidate = null;
for(int num :nums){
if(count == 0) {
candidate = num;
}
count += (num == candidate) ? 1 : -1;
}
return candidate;
}
}
随机切分
在数组中随机选取一个元素作为切分元素进行切分,并返回下标。由于含有多数元素的数组排序后数组的中间元素就为多数元素,所以
- 如果切分元素下标等于num.length/2则该元素就为多数元素;
- 如果切分元素下标小于num.length/2则对该元素右边继续切分;
- 如果切分元素下标大于num.length/2则对该元素左边继续切分;
class Solution {
public int majorityElement(int[] nums) {
if(nums == null || nums.length == 0) return 0;
int lo = 0;
int hi = nums.length - 1;
int middle = nums.length >> 1;
int index = partition(nums, lo, hi);
while(index != middle){
if(index > middle){
hi = index - 1;
index = partition(nums, lo, hi);
}else if(index < middle){
lo = index + 1;
index = partition(nums, lo, hi);
}
}
return nums[middle];
}
/**
* 随机切分函数
*/
private int partition(int[] a, int lo, int hi){
if(a == null || a.length == 0 || lo < 0 || hi > a.length)
throw new RuntimeException("Invalid Parameters");
int index = randRange(new Random(), lo, hi+1);
exch(a, hi, index);
int small = lo - 1;
for(index = lo; index < hi; ++index){
if(a[index] < a[hi] ){
++small;
if(index != small){
exch(a, small, index);
}
}
}
small++;
exch(a, small, hi);
return small;
}
private int randRange(Random rand, int min, int max) {
if(max - min == 0) return min;
int r = rand.nextInt(max - min) + min;
return r;
}
private void exch(int[] nums, int i, int j){
int temp = nums[i];
nums[i] = nums[j];
nums[j] = temp;
}
}