There are two sorted arrays nums1 and nums2 of size m and n respectively.
Find the median of the two sorted arrays. The overall run time complexity should be O(log (m+n)).
You may assume nums1 and nums2 cannot be both empty.
Example 1:
nums1 = [1, 3]
nums2 = [2] The median is 2.0
Example 2:
nums1 = [1, 2]
nums2 = [3, 4] The median is (2 + 3)/2 = 2.5
Solution1: time complexity is O(m+n). merge two sorted array, then find the median
code:
/*
Time Complexity: O(m+n)
Space Complexity: O(m+n)
*/ class Solution {
public double findMedianSortedArrays(int[] nums1, int[] nums2) { // find median
int[] mergedArray = mergeTwoSortedArray(nums1, nums2);
int n = mergedArray.length;
if( n % 2 == 0){
return (mergedArray[(n-1)/2] + mergedArray[n/2])/2.0;
}else{
return mergedArray[n/2];
}
} public int[] mergeTwoSortedArray(int[] nums1, int[] nums2){ // merge sort
int[] merged = new int[nums1.length + nums2.length];
int i = 0, j = 0, k = 0;
// 两个array同时扫
while( i < nums1.length && j < nums2.length){
merged[k++] = (nums1[i] < nums2[j]) ? nums1[i++] : nums2[j++];
}
//扫到只剩下较长的那个array， either nums1 or nums2
while ( i < nums1.length ){
merged[k++] = nums1[i++];
}
while(j < nums2.length){
merged[k++] = nums2[j++];
}
return merged;
}
}
Solution2: time complexity is O(log (m+n)).
这个题的思路可以是找出2个sorted array 所有元素中，Kth 元素
因此我们可以把问题转化成，
“Given 2 sorted arrays, A, B of sizes m, n respectively, find the numbers which are NOT medians of the two sorted arrays”
如果我们能找到那些NOT medians的数字，删掉，并不断缩小范围，那最终剩下的一定就是actual median
Step1, 确定要找的median是merged之后array中的第几个元素
Step2, 用binary search切开nums1, nums2。发现nums1左半边的长度为4， nums2左半边的长度为2。
Step3, nums1左半边长度+ nums2左半边长度为6
Step4， 所以第7个元素可能在nums1的右半边pivot上，或者在nums2的右半边pivot上
比较发现，第7个元素应该是7
code：
/*
Time Complexity: O(log(m+n))
Space Complexity: O(log(m+n))
*/ class Solution {
public double findMedianSortedArrays(final int[] A, final int[] B) {
int total = A.length + B.length;
if (total %2 == 0){
return (findKth(A, 0, B, 0, total / 2) + findKth(A, 0, B, 0, total / 2 + 1)) / 2.0;
}else{
return findKth(A, 0, B, 0, total / 2 + 1);
}
} private static int findKth(final int[] A, int ai, final int[] B, int bi, int k) {
//always assume that A is shorter than B
if (A.length - ai > B.length - bi) {
return findKth(B, bi, A, ai, k);
}
if (A.length - ai == 0) return B[bi + k - 1];
if (k == 1) return Math.min(A[ai], B[bi]); //divide k into two parts
int k1 = Math.min(k / 2, A.length - ai), k2 = k - k1;
if (A[ai + k1 - 1] < B[bi + k2 - 1])
return findKth(A, ai + k1, B, bi, k - k1);
else if (A[ai + k1 - 1] > B[bi + k2 - 1])
return findKth(A, ai, B, bi + k2, k - k2);
else
return A[ai + k1 - 1];
}
}