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| 1 | +想法:先選定較小的矩陣,然後用 binary search 找到合適的 partitionA 以及 partitionB |
| 2 | +使之符合 partitionA + partitionB = (m + n + 1) / 2 |
| 3 | +切出 nums1[0 .. partitionA - 1] , nums2[0 .. partitionB - 1] 以及 nums1[partitionA .. m] , nums2[partitionB .. n] |
| 4 | +且 nums1[partitionA - 1] <= nums2[partitionB] and nums2[partitionB - 1] <= nums1[partitionA] |
| 5 | +這樣就代表剛好切到中線上,反之則依照大小關係向左向右推進 |
| 6 | + |
| 7 | +Time Complexity : O( log(min(m , n))) for binary searching on the smaller array |
| 8 | +Space Complexity : O(1) for variables |
| 9 | + |
| 10 | +class Solution { |
| 11 | +public: |
| 12 | + double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) { |
| 13 | + if ( nums1.size() > nums2.size() ) |
| 14 | + swap(nums1 , nums2 ) ; |
| 15 | + |
| 16 | + int m = nums1.size() , n = nums2.size() ; |
| 17 | + int l = 0 , r = m ; |
| 18 | + while (l <= r) { |
| 19 | + int partitionA = (l + r) / 2 ; |
| 20 | + int partitionB = (m + n + 1) / 2 - partitionA ; |
| 21 | + |
| 22 | + int maxleftA = (partitionA == 0)? INT_MIN : nums1[partitionA - 1] ; |
| 23 | + int minrightA = (partitionA == m)? INT_MAX : nums1[partitionA] ; |
| 24 | + int maxleftB = (partitionB == 0)? INT_MIN : nums2[partitionB - 1] ; |
| 25 | + int minrightB = (partitionB == n)? INT_MAX : nums2[partitionB] ; |
| 26 | + |
| 27 | + if ( maxleftA <= minrightB && maxleftB <= minrightA ) { |
| 28 | + // the cut is right |
| 29 | + if ( (m + n) % 2 == 0 ) { |
| 30 | + return (max(maxleftA,maxleftB)+min(minrightA,minrightB)) / 2.0; |
| 31 | + } |
| 32 | + else |
| 33 | + return max(maxleftA , maxleftB) ; |
| 34 | + } |
| 35 | + else if ( maxleftA > minrightB ) |
| 36 | + r = partitionA - 1 ; |
| 37 | + else |
| 38 | + l = partitionA + 1 ; |
| 39 | + } |
| 40 | + return 0 ; |
| 41 | + } |
| 42 | +}; |
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