leetcode-lintcode/004_median_of_two_sorted_arrays.java at master · lt616/leetcode-lintcode · GitHub

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/* Solution 01: Brute force O(m + n) */
class Solution {
    public double findMedianSortedArrays(int[] nums1, int[] nums2) {
        int size = nums1.length + nums2.length;
        boolean isOdd = (size % 2 == 1);
        int medianSum = 0;
        int m = 0;
        int n = 0;
        for (int i = 0; i < size; i++) {
            int current;
            if (m >= nums1.length || (n < nums2.length && nums1[m] >= nums2[n])) {
                current = nums2[n];
                n ++;
            } else {
                current = nums1[m];
                m ++;
            }

            if (isOdd) {
                if (i == size / 2) {
                    medianSum += current;
                }
            } else {
                if (i == size / 2 || i == size / 2 - 1) {
                    medianSum += current;
                }
            }
        }

        double median = (double) medianSum;
        System.out.println(median);
        return (isOdd) ? median : median / 2;
    }
}

/* Solution 02: Binary search O(log(min(m, n))) */
class Solution {
    public double findMedianSortedArrays(int[] nums1, int[] nums2) {
        if (nums1.length == 0 && nums2.length == 0)
            return 0;

        if (nums1.length == 0)
            return findMedianSingleArray(nums2);

        if (nums2.length == 0)
            return findMedianSingleArray(nums1);

        int start = 0;
        int end = nums1.length;
        int size = nums1.length + nums2.length;
        while (start + 1 < end) {
            int i = (start + end) / 2;
            int j = size / 2 - i;
            if (j < 0) {
                end = i;
                continue;
            }

            if (j >= nums2.length) {
                start = i;
                continue;
            }


            if (isMedian(nums1, nums2, i, j, size)) {
                return getMedian(nums1, nums2, i, j, size);
            } else {
                if (findLargerValue(nums1, nums2, i, j)) {
                    end = i;
                } else {
                    start = i;
                }
            }
        }

        /* if start is median */
        int i = start;
        int j = size / 2 - i;
        if (j >= 0) {
            if (isMedian(nums1, nums2, i, j, size))
                return getMedian(nums1, nums2, i, j, size);
        }

        /* if end is median */
        i = end;
        j = size / 2 - i;
        if (j >= 0) {
            if (isMedian(nums1, nums2, i, j, size))
                return getMedian(nums1, nums2, i, j, size);
        }

        return 0;
    }

    private boolean isMedian(int[] nums1, int[] nums2, int i, int j, int size) {
        int maxLeft = findValue(nums1, i - 1, nums2, j - 1, true);
        int minRight = findValue(nums1, i, nums2, j, false);
        return (maxLeft <= minRight);
    }

    private double getMedian(int[] nums1, int[] nums2, int i, int j, int size) {
        int maxLeft = findValue(nums1, i - 1, nums2, j - 1, true);
        int minRight = findValue(nums1, i, nums2, j, false);

        if (size % 2 == 1) {
            return (double) minRight;
        } else {
            return ((double) maxLeft + minRight) / 2;
        }
    }

    private double findMedianSingleArray(int[] nums) {
        int mid = nums.length / 2;
        if (nums.length % 2 == 1) {
            return (double) nums[mid];
        } else {
            return ((double) nums[mid - 1] + nums[mid]) / 2;
        }
    }

    private int findValue(int[] nums1, int i, int[] nums2, int j, boolean isMax) {
        if (i < 0 || i >= nums1.length)
            return nums2[j];

        if (j < 0 || j >= nums2.length)
            return nums1[i];

        if (isMax)
            return (nums1[i] > nums2[j]) ? nums1[i] : nums2[j];
        else
            return (nums1[i] < nums2[j]) ? nums1[i] : nums2[j];
    }

    private boolean findLargerValue(int[] nums1, int[] nums2, int i, int j) {
        if (i < 0 || i >= nums1.length)
            return false;

        if (j < 0 || j >= nums2.length)
            return true;

        return (nums1[i] >= nums2[j]);
    }
}

/* Official solution 02 */
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;
    }
}