[leetCode刷题笔记]2017.03.24

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274. H-Index

这道题用一个新的数组，存储每个每个引用次数出现的频率，注意如果某文章引用次数大于科学家发文章的总数，则看作等于发文章总数的频率。

然后再从前往后加计算满足H-INDEX条件的最大值。

public class Solution {    public int hIndex(int[] citations) {        int len = citations.length;        if (len <=0) {            return 0;        }        int[] arr= new int[len + 1];        for (int i = 0; i < len; i++) {            if (citations[i] >= len) {                arr[len]++;            }            else {                arr[citations[i]]++;            }        }        int k = 0;        int res = 0;        for (int i = len; i >= 0; i--) {            k += arr[i];            if (k >= i) {                return i;            }        }        return 0;    }}

324. Wiggle Sort II

先找出数组中中位数，然后小于这个数的放奇数，其他放偶数。找中位数用到快排思想。

public class Solution {    public void wiggleSort(int[] nums) {        int median = findKthLargest(nums, (nums.length  + 1) / 2);        int n = nums.length;                int left = 0, i = 0, right = n - 1;                while (i <= right) {            if (nums[newIndex(i,n)] > median) {                swap(nums, newIndex(left++,n), newIndex(i++,n));            }            else if (nums[newIndex(i,n)] < median) {                swap(nums, newIndex(right--,n), newIndex(i,n));            }            else {                i++;            }        }    }    private int newIndex(int index, int n) {        return (1 + 2 * index) % (n | 1);    }    private int findKthLargest(int[] nums, int k) {        return findKthLargest(nums, k, 0, nums.length - 1);    }    private int findKthLargest(int[] nums, int k, int l, int r) {        if (l >= r) {            return nums[l];        }        int m = partition(nums, l, r);        if (m == k) {            return nums[m];        }        else if (m < k) {            return findKthLargest(nums, k, m + 1, r);        }        else {            return findKthLargest(nums, k, l, m - 1);        }            }    private int partition(int[] nums, int l, int r) {        int pivot = nums[l];          int m = l;          int n = l+1;          while(n <= r) {              if(nums[n] < pivot){                  swap(nums, ++m, n);              }              n++;          }          swap(nums, l, m);          return m;    }    public void swap(int[] nums, int a, int b) {          int temp = nums[a];          nums[a] = nums[b];          nums[b] = temp;      }}

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