Created
February 18, 2020 08:33
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Median of Two Sorted Arrays
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function medianOfTwoSorted(sortedA: Array<number>, sortedB: Array<number>) { | |
let [m, n] = [sortedA.length, sortedB.length]; | |
if (m > n) { | |
// Use the longest array for m. | |
[sortedA, sortedB, m, n] = [sortedB, sortedA, n, m]; | |
} | |
if (n === 0) { | |
throw new Error('Array should not be empty'); | |
} | |
let [iMin, iMax, halfLength] = [0, m, Math.floor((m + n + 1) / 2)]; | |
while (iMin <= iMax) { | |
let i = Math.floor((iMin + iMax) / 2); | |
let j = halfLength - i; | |
if (i < m && sortedB[j - 1] > sortedA[i]) { | |
iMin = i + 1; | |
continue; | |
} | |
if (i > 0 && sortedA[i - 1] > sortedB[j]) { | |
iMax = i - 1; | |
continue; | |
} | |
let maxOfLeft = | |
i === 0 | |
? sortedB[j - 1] | |
: j === 0 | |
? sortedA[i - 1] | |
: Math.max(sortedA[i - 1], sortedB[j - 1]); | |
if ((m + n) % 2 === 1) { | |
return maxOfLeft; | |
} | |
let minOfRight = | |
i === m | |
? sortedB[j] | |
: j === n | |
? sortedA[i] | |
: Math.min(sortedA[i], sortedB[j]); | |
return (maxOfLeft + minOfRight) / 2; | |
} | |
} |
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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 | |
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