問題描述
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
在我的一篇博客中提供了類似的方法找到第K個最小數(shù)
這里提供另外一種想法
想法
把數(shù)組a b 都分為兩個數(shù)組 恩尾，a分為a[0] - a[i-1] , a[i] - a[m-1];b分為b[0] - b[j-1],b[j] - b[n-1]；
滿足下列兩個條件
- i + j = m - i + n - j 也就是 j = (m+n) / 2 - j;
- 左邊的小于右邊的 也就是a[i-1] < b[j]; b[j-1] < a[i]
滿足這兩個條件之后，很容易得到兩個數(shù)組得中值 柿隙，也就是如果數(shù)組是奇數(shù)的話垒玲，就是a[i-1],b[j-1]較大的那個纳猫，如果是偶數(shù)恕出，就是兩者相加除以2.
問題就轉換到了如何分成滿足這兩個條件的數(shù)組蔼卡，也就是找到合適的i,也就是找到滿足第二個條件的i
找i方法
找[imin,imax]
- 如果a[i-1] > b[j]屯伞，就要減少也就是imax = i - 1; 找[imin, i-1]腿箩；
- 如果a[i] < b[j-1] i要增大 就是找[i+1,imax]，也就是imin = i + 1;
代碼
class Solution {
public double findMedianSortedArrays(int[] A, int[] B) {
int m = A.length;
int n = B.length;
if (m > n) { // to ensure m<=n
int[] temp = A; A = B; B = temp;
int tmp = m; m = n; n = tmp;
}
int iMin = 0, iMax = m, halfLen = (m + n + 1) / 2;
while (iMin <= iMax) {
int i = (iMin + iMax) / 2;
int j = halfLen - i;
if (i < iMax && B[j-1] > A[i]){
iMin = i + 1; // i is too small
}
else if (i > iMin && A[i-1] > B[j]) {
iMax = i - 1; // i is too big
}
else { // i is perfect
int maxLeft = 0;
if (i == 0) { maxLeft = B[j-1]; }
else if (j == 0) { maxLeft = A[i-1]; }
else { maxLeft = Math.max(A[i-1], B[j-1]); }
if ( (m + n) % 2 == 1 ) { return maxLeft; }
int minRight = 0;
if (i == m) { minRight = B[j]; }
else if (j == n) { minRight = A[i]; }
else { minRight = Math.min(B[j], A[i]); }
return (maxLeft + minRight) / 2.0;
}
}
return 0.0;
}
}
