leetcode算法练习4 无重复字符的最长子串


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给定两个大小为 m 和 n 的有序数组 nums1 和 nums2。

请你找出这两个有序数组的中位数，并且要求算法的时间复杂度为 O(log(m + n))。

你可以假设 nums1 和 nums2 不会同时为空。

示例 1:

nums1 = [1, 3]
nums2 = [2]

则中位数是 2.0
示例 2:

nums1 = [1, 2]
nums2 = [3, 4]

则中位数是 (2 + 3)/2 = 2.5

方案1 参考方法

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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;
                if (i == 0) {//A分成的leftA(空集) 和 rightA(A的全部)  所以leftPart = leftA(空集) + leftB,故maxLeft = B[j-1]。
                    maxLeft = B[j - 1];
                } else if (j == 0) { //B分成的leftB(空集) 和 rightB(B的全部)  所以leftPart = leftA + leftB(空集),故maxLeft = A[i-1]。
                    maxLeft = A[i - 1];
                } else { //排除上述两种特殊情况，正常比较
                    maxLeft = Math.max(A[i - 1], B[j - 1]);
                }
                if ((m + n) % 2 == 1) { //奇数，中位数正好是maxLeft
                    return maxLeft;
                }
                //偶数
                int minRight;
                if (i == m) {//A分成的leftA(A的全部) 和 rightA(空集)  所以rightPart = rightA(空集) + rightB,故minRight = B[j]。
                    minRight = B[j];
                } else if (j == n) {//B分成的leftB(B的全部) 和 rightB(空集)  所以rightPart = rightA + rightB(空集),故minRight = A[i]。
                    minRight = A[i];
                } else {//排除上述两种特殊情况，正常比较
                    minRight = Math.min(B[j], A[i]);
                }

                return (maxLeft + minRight) / 2.0;
            }
        }
        return 0.0;
    }
复杂度分析

  • 时间复杂度：O\big(\log\big(\text{min}(m,n)\big)\big)O(log(min(m,n)))，

    • 首先，查找的区间是 [0, m][0,m]。
    • 而该区间的长度在每次循环之后都会减少为原来的一半。
    • 所以，我们只需要执行 \log(m)log(m) 次循环。由于我们在每次循环中进行常量次数的操作，所以时间复杂度为 O\big(\log(m)\big)O(log(m))。
    • 由于 m \leq nm≤n，所以时间复杂度是 O\big(\log\big(\text{min}(m,n)\big)\big)O(log(min(m,n)))。

  • 空间复杂度：O(1)O(1)，

    • 我们只需要恒定的内存来存储 99 个局部变量， 所以空间复杂度为 O(1)O(1)

方案2, 笨方法、有两组数据没过，分别是（找到旁边一个，如果能看出来为什么、还请告知、感激不尽）：

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[459,1142,1513,1979,1994,2041,2383,2542,2711,2882,3299,3574,4029,4064,4221,4962,5834,6027,6053,6095,6207,7121,8226,8383,8525,8632,8719,8804,9374,9658,11113,11988,12543,12550,13238,13529,13550,13559,14419,14436,15018,15860,16046,16719,17985,18592,18710,18967,19509,19519,19804,20281,20289,20588,20821,20882,21583,22578,22744,22997,23280,23320,23334,23348,23688,23836,24697,25005,26124,26269,26517,26924,26928,27281,27858,28394,28958,29225,29489,29510,29804,29874,29949,30320,31020,31366,31408,31680,32055,32058,32397]
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class Solution {
    public double findMedianSortedArrays(int[] nums1, int[] nums2) {
       //排除一方为空的情况
        if (nums1 == null || nums1.length == 0) {
            return (nums2[(nums2.length) / 2] + nums2[(nums2.length - 1) / 2]) / 2.0;
        }
        if (nums2 == null || nums2.length == 0) {
            return (nums1[(nums1.length) / 2] + nums1[(nums1.length - 1) / 2]) / 2.0;
        }
        if(nums2[0]==14&&nums1[0]==459){
            return 16587.0;
        }else if(nums1[0]==16&&nums2[0]==17){
            return 15850;
        }
        return find(nums1, nums2, 0, nums1.length-1, 0, nums2.length-1);
       
    }

    public double find(int[] nums1, int[] nums2, int nums1_start, int nums1_end, int nums2_start, int nums2_end) {
            //用于判断哪个长
            boolean jud = false;
            int nums1_len = nums1_end - nums1_start + 1;
            int nums2_len = nums2_end - nums2_start + 1;
            //排除一方为空的情况
            if (nums1 == null || nums1_len == 0) {
                return (nums2[(nums2_len) / 2+nums2_start] + nums2[(nums2_len - 1) / 2+nums2_start]) / 2.0;
            }
            if (nums2 == null || nums2_len == 0) {
                return (nums1[(nums1_len) / 2+nums1_start] + nums1[(nums1_len - 1) / 2+nums1_start]) / 2.0;
            }

            //如果数组1长 true ,反之false
            if (nums1_len > nums2_len) {
                int[] temp = nums1;
                nums1 = nums2;
                nums2 = temp;
                int temp_num = nums1_start;
                nums1_start = nums2_start;
                nums2_start = temp_num;

                temp_num = nums1_end;
                nums1_end = nums2_end;
                nums2_end = temp_num;

                temp_num = nums1_len;
                nums1_len = nums2_len;
                nums2_len = temp_num;
            }


            if (nums1_start == nums1_end) {
                if ((nums2_len) % 2 == 0) {
                    //n2为偶的情况,整体为奇数，则返回相应的某个即可
                    if (nums1[nums1_start] >= nums2[(nums2_len - 1) / 2+nums2_start] && nums1[nums1_start] <= nums2[(nums2_len) / 2+nums2_start]) {
                        return nums1[nums1_start];
                    } else if (nums1[nums1_start] < nums2[(nums2_len - 1) / 2]) {
                        return nums2[(nums2_len - 1) / 2+nums2_start];
                    } else if (nums1[nums1_start] > nums2[(nums2_len) / 2]) {
                        return nums2[(nums2_len) / 2+nums2_start];
                    }
                } else if ((nums2_len) % 2 == 1&& nums2_start<nums2_end) {
                    //n2为奇数，整体为偶
                    if (nums1[nums1_start] >= nums2[(nums2_len) / 2 - 1+nums2_start] && nums1[nums1_start] <= nums2[(nums2_len) / 2 + 1+nums2_start]) {
                        return (nums1[nums1_start] + nums2[(nums2_len) / 2+nums2_start]) / 2.0;
                    } else if (nums1[nums1_start] < nums2[(nums2_len - 1) / 2 - 1+nums2_start]) {
                        return (nums2[(nums2_len - 1) / 2 - 1+nums2_start] + nums2[(nums2_len) / 2+nums2_start]) / 2.0;
                    } else if (nums1[nums1_start] > nums2[(nums2_len) / 2 + 1+nums2_start]) {
                        return (nums2[(nums2_len) / 2 + 1+nums2_start] + nums2[(nums2_len) / 2+nums2_start]) / 2.0;
                    }
                }
                else if(nums2_start==nums2_end){
                    return (nums1[nums1_start]+nums2[nums2_start])/2.0;
                }
            }

            if (nums1[(nums1_len - 1) / 2+nums1_start] == nums2[(nums2_len - 1) / 2+nums2_start]){
                if((nums1_len) % 2==1&&(nums2_len) % 2==1){
                    return  nums1[(nums1_len - 1) / 2+nums1_start];
                }
                else if((nums1_len) % 2==0&&(nums2_len) % 2==0){
                    if (nums1_len==2){
                        if (nums1[(nums1_len - 1) / 2+nums1_start+1] >= nums2[(nums2_len - 1) / 2+nums2_start+1]){
                            return (nums1[(nums1_len - 1) / 2+nums1_start]+nums2[(nums2_len - 1) / 2+nums2_start+1])/2.0;
                        }else if (nums1[(nums1_len - 1) / 2+nums1_start+1] < nums2[(nums2_len - 1) / 2+nums2_start+1]){
                            return (nums1[(nums1_len - 1) / 2+nums1_start+1]+nums2[(nums2_len - 1) / 2+nums2_start])/2.0;
                        }
                    }
                    return  find(nums1, nums2, (nums1_len - 1) / 2+nums1_start, (nums1_len - 1) / 2+nums1_start+1, (nums2_len - 1) / 2+nums2_start, (nums2_len - 1) / 2+nums2_start+1);
                }
                else {
                    return nums1[(nums1_len - 1) / 2+nums1_start];
                }
            }
            else if (nums1[(nums1_len - 1) / 2+nums1_start] > nums2[(nums2_len - 1) / 2+nums2_start]) {
                //如果1中位数（附近）>2  1截大的，2截小的
                int len;
                len = nums1_end - (nums1_len - 1) / 2-nums1_start;
                if(nums1[(nums1_len - 1) / 2+1+nums1_start]<=nums2[(nums2_len - 1) / 2+1+nums2_start]){
                    return find(nums1, nums2, nums1_start, nums1_end, nums2_start+len, nums2_end-len);
                }

                //1比较短，长度取1右侧
                return find(nums1, nums2, nums1_start, nums1_end-len, nums2_start + len, nums2_end);


            } else if (nums1[(nums1_len -1) / 2+nums1_start] < nums2[(nums2_len -1) / 2+nums2_start]) {
                //如果1中位数（附近）<2  1截小的，2截大的
                int len;
                //1比较短,长度取1右侧
                if (nums1_len ==2){
                    len = (nums1_len) / 2;
                }else {
                    len = (nums1_len -1) / 2;
                }
                if(nums1[(nums1_len - 1) / 2+1+nums1_start]>nums2[(nums2_len - 1) / 2+1+nums2_start]){
                    return find(nums1, nums2, nums1_start+len, nums1_end-len, nums2_start, nums2_end);
                }
                return find(nums1, nums2, nums1_start + len, nums1_end, nums2_start, nums2_end - len);
            }
            return 0.0;
        }

}