【群视频】笔记 - 2015.06.10

【题目1】

public int getLessIndex(int[] arr) {    if (arr == null || arr.length == 0) {        return -1; // no exist    }    if (arr.length == 1 || arr[0] < arr[1]) {         return 0;    }    if (arr[arr.length - 1] < arr[arr.length - 2]) {        return arr.length - 1;    }    int left = 1;    int right = arr.length - 2;    int mid = 0;    while (left < right) {        mid = (left + right) / 2;        if (arr[mid] > arr[mid - 1]) {            right = mid - 1;        } else if (arr[mid] > arr[mid + 1]) {            left = mid + 1;        } else {            return mid;        }    }    return left;}

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【题目2】

时间 O(n)
额外空间O(1)

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【题目3】

时间复杂度O(h^2) = [log(n)]^2

public int nodeNum(Node head) {    if (head == null) {        return 0;    }    return bs(head, 1, mostLeftLevel(head, 1));}public int bs(Node node, int l, int h) {    if (l == h) {        return 1;    }    if (mostLeftLevel(node.right, l + 1) == h) {        return (1 << (h - l)) + bs(node.right, l + 1, h);    } else {        return (1 << (h - l - 1)) + bs(node.left, l + 1, h);    }}public int mostLeftLevel(Node node, int level) {    while (node != null) {        level++;        node = node.left;    }    return level - 1;}

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【题目4】

public static int getUpMedian(int[] arr1, int[] arr2) {    if (arr1 == null || arr2 == null || arr1.length != arr2.length) {        throw new RuntimeException("Your arr is invalid!");    }    return findProcess(arr1, 0, arr1.length - 1, arr2, 0, arr2.length - 1);}//arr1的 start1~end1范围上 、arr2的 start2~end2范围上public static int findProcess(int[] arr1, int start1, int end1, int[] arr2, int start2, int end2) {    if (start1 == end1) {        return Math.min(arr1[start1], arr2[start2]);    }    // 元素个数为奇数，则offset为0；元素个数为偶数，则offset为1；    int offset = ((end1 - start1 + 1) & 1) ^ 1;    int mid1 = (start1 + end1) / 2;    int mid2 = (start2 + end2) / 2;    if (arr1[mid1] > arr2[mid2]) {        return findProcess(arr1, start1, mid1, arr2, mid2 + offset, end2);    } else if (arr1[mid1] < arr2[mid2]) {        return findProcess(arr1, mid1 + offset, end1, arr2, start2, mid2);    } else {        //arr1[mid1] = arr2[mid2]        return arr1[mid1];    }}

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【题目5】

Ⅰ. arr1拿出k个数，arr2拿出k个数，求arr1和arr2的上中位数

Ⅱ.

Ⅲ .找 1’~27’ 该淘汰哪一段

public static int findKthNum(int[] arr1, int[] arr2, int kth) {    if (arr1 == null || arr2 == null) {        throw new RuntimeException("Your arr is invalid!");    }    if (kth < 1 || kth > arr1.length + arr2.length) {        throw new RuntimeException("K is invalid!");    }    int[] longArr = arr1.length >= arr2.length ? arr1 : arr2;    int[] shortArr = arr1.length < arr2.length ? arr1 : arr2;    int lenL = longArr.length;    int lenS = shortArr.length;    if (kth <= lenS) {        return getUpMedian(shortArr, 0, kth - 1, longArr, 0, kth - 1);    }    if (kth > lenL) {        if (shortArr[kth - lenL - 1] >= longArr[lenL - 1]) {            return shortArr[kth - lenL - 1];        }        if (longArr[kth - lenS - 1] >= shortArr[lenS - 1]) {            return longArr[kth - lenS - 1];        }        return getUpMedian(shortArr, kth - lenL, lenS - 1, longArr, kth - lenS, lenL - 1);    }    if (longArr[kth - lenS - 1] >= shortArr[lenS - 1]) {        return longArr[kth - lenS - 1];    }    return getUpMedian(shortArr, 0, lenS - 1, longArr, kth - lenS, kth - 1);}public static int getUpMedian(int[] arr1, int start1, int end1, int[] arr2,    int start2, int end2) {        if (start1 == end1) {            return Math.min(arr1[start1], arr2[start2]);        }    int offset = ((end1 - start1 + 1) & 1) ^ 1;    int mid1 = (start1 + end1) / 2;    int mid2 = (start2 + end2) / 2;    if (arr1[mid1] > arr2[mid2]) {        return getUpMedian(arr1, start1, mid1, arr2, mid2 + offset, end2);    } else if (arr1[mid1] < arr2[mid2]) {        return getUpMedian(arr1, mid1 + offset, end1, arr2, start2, mid2);    } else {        return arr1[mid1];    }}