Interview questions 17.10. Main elements
Title description
Elements that account for more than half of the array are called main elements. Given an integer array, find its main element. If not, return -1.
Example 1:
Input: [1,2,5,9,5,9,5,5,5]
Output: 5
Example 2:
Input: [3,2]
Output: -1
Example 3:
Input: [2,2,1,1,1,2,2]
Output: 2
Problem-solving ideas
There are several methods involved, it is recommended to verify the intermediate elements after sorting
import java.util.Arrays;
import java.util.HashMap;
import java.util.Map;
//leetcode submit region begin(Prohibit modification and deletion)
class Solution_17_10 {
public int majorityElement_1(int[] nums) {
/**
* 哈希表 ，时间复杂度 O（N2），空间复杂度O（N）
*/
HashMap<Integer,Integer> map = new HashMap<>();
int p = nums.length / 2;
for (int i = 0; i < nums.length; i++) {
map.put(nums[i],map.getOrDefault(nums[i],0) + 1);
}
for (Map.Entry<Integer, Integer> entry : map.entrySet()) {
if(entry.getValue() > p){
return entry.getKey();
}
}
return -1;
}
/**
* 排序后，使用双指针，统计大于超过一半长度的元素
* @param nums
* @return
*/
public int majorityElement_2(int[] nums) {
if (nums.length == 1){
return nums[0];
}
Arrays.sort(nums);
System.out.println(Arrays.toString(nums));
int start = 0;
int end = 0;
for (int i = 1; i < nums.length; i++) {
if (nums[i - 1] == nums[i]){
end++;
}else {
start = end + 1;
end++;
}
if (end - start + 1 > nums.length / 2){
return nums[start];
}
}
return -1;
}
/**
* 排序后验证中间元素
* @param nums
* @return
*/
public int majorityElement_3(int[] nums) {
Arrays.sort(nums);
int ans = nums[nums.length / 2];
int number = 0;
for (int num : nums) {
if (num == ans){
number++;
}
}
if (number > nums.length / 2){
return ans;
}
return -1;
}
/**
* 首先给这个数组排序，然后如果有主要元素，那么这个元素一定在中间，
* 可以定义两个指针，从这个数的左右开始扫描，扫描到一个相同的数。
* 令左指针减一或右指针加一，当扫描完成后，令右指针-左指针-2+1，
* 由于一开始两个指针相差2，所以要减去2，这样的话就算得了扫描的个数，
* 但是还有中间的那个数没有算，所以需要加一，最后判断这个算得的数是否大于数组的一半即可
* 时间复杂度：排序的时候是快排O(n*logn)+O(n)+O(n)，空间为O(1)
*
*/
public int majorityElement(int[] nums) {
Arrays.sort(nums);
int left = nums.length/2-1, right = nums.length/2+1, num = nums[nums.length/2];
while (left >= 0 && nums[left] == num) {
left --;
}
while (right < nums.length && nums[right] == num) {
right ++;
}
if (right - left - 1> nums.length/2) {
return num;
}
return -1;
}
/**
* 摩尔投票法
* @param nums
* @return
*/
public int majorityElement_4(int[] nums) {
int temp = 0,count = 0;
for (int num : nums) {
if (count == 0){
temp = num;
count = 1;
}
else {
if (temp == num){
count++;
}else {
count--;
}
}
}
if (count > 0){
int n = 0;
for (int i : nums) {
if (i == temp)
n++;
}
if (n > nums.length / 2){
return temp;
}
}
return -1;
}
}
//leetcode submit region end(Prohibit modification and deletion)