找两个有序数组的中位数的几种方式

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leetcode中第4题 ‘找两个有序数组的中位数’的3种方式时间复杂度分别是O(m+n) ,O(k),O(log(m+n))

    /**1     * O(m+n)     * 合并两有序数组成一个新有序数组，再按中间位置取值     * @param nums1     * @param nums2     * @return     */    public static double findMedianSortedArrays1(int[] nums1, int[] nums2) {        int length1 = nums1.length;        int length2 = nums2.length;        int lengthall = length1 + length2;        int[] all = new int[length1+length2];        int i=0,j=0,k=0;        while (i<length1 && j <length2){            if(nums1[i]<nums2[j]){                all[k]=nums1[i];                i++;                k++;            }else{                all[k] = nums2[j];                j++;                k++;            }        }        while (i<length1){            all[k]=nums1[i];            i++;            k++;        }        while (j<length2){            all[k] = nums2[j];            j++;            k++;        }        double result;        if (lengthall % 2 == 0){            result = ( all[lengthall/2-1] + all[lengthall/2] )*1.0/2;        }else{            result = all[lengthall/2];        }        return result;    }    /**2     * O(k)     * 两个指针分别从两数组开头指，比较两指针处的数，谁小谁往后移，     * 直到两指针共移动k次，k为中间位置     * @param nums1     * @param nums2     * @return     */    public static double findMedianSortedArrays2(int[] nums1, int[] nums2) {        int length1 = nums1.length;        int length2 = nums2.length;        int lengthall = length1 + length2;        int inx1 = 0;        int inx2 = 0;        int x = -1;        int y = -1;        while (inx1 + inx2 < lengthall){            if (inx1 < length1) {                while (inx2 == length2 ||  nums1[inx1] <= nums2[inx2] ) {                    inx1++;                    if (inx1 + inx2 == (lengthall + 1) / 2) {                        x = nums1[inx1 - 1];                    }                    if (inx1 + inx2 == (lengthall + 2) / 2) {                        y = nums1[inx1 - 1];                        return (x + y) * 1.0 / 2;                    }                    if (inx1 == length1){                        break;                    }                }            }            if (inx2 < length2){                while ( inx1 == length1 || nums2[inx2] <= nums1[inx1] ) {                    inx2++;                    if (inx1 + inx2 == (lengthall + 1)/2){                        x = nums2[inx2-1];                    }                    if (inx1 + inx2 == (lengthall + 2)/2){                        y = nums2[inx2-1];                        return (x+y)*1.0/2;                    }                    if (inx2 == length2){                        break;                    }                }            }        }        return -1;    }    /**3     * O(log(m+n))     * 两个数组分别采用二分法查找     * @param nums1     * @param nums2     * @return     */    public static double findMedianSortedArrays3(int[] nums1, int[] nums2) {        int length1 = nums1.length;        int length2 = nums2.length;        int lengthall = length1 + length2;        int l = (lengthall+1)/2;        int r = (lengthall+2)/2;        return (getKMin(nums1,0,nums2,0,l)+getKMin(nums1,0,nums2,0,r))*1.0/2;    }    public static int getKMin(int[] A, int Astart, int[] B, int Bstart, int k){        if (Astart>A.length-1){            return B[Bstart + k -1];        }        if (Bstart>B.length-1){            return A[Astart + k -1];        }        if (k==1){            return Math.min(A[Astart],B[Bstart]);        }        int Amin = Integer.MAX_VALUE,Bmin = Integer.MAX_VALUE;        if (Astart + k/2 -1 < A.length){            Amin = A[Astart + k/2 -1];        }        if (Bstart + k/2 -1 < B.length){            Bmin = B[Bstart + k/2 -1];        }        return Amin < Bmin ? getKMin(A,Astart + k/2, B, Bstart,k-k/2):getKMin(A,Astart,B,Bstart+k/2,k-k/2);    }

public static void main(String[] args) {        int NUM = 5000000;        int[] nums1 = new int[NUM];        int[] nums2 = new int[NUM];        nums1[0] = 1;        nums2[0] = 4;        for (int i=1;i<NUM;i++){            Random random = new Random();            nums1[i] = nums1[i-1] + random.nextInt(5);            nums2[i] = nums2[i-1] + random.nextInt(5);        }        long start = System.currentTimeMillis();        double result1 = findMedianSortedArrays1(nums1, nums2);        long end = System.currentTimeMillis();        System.out.println("result 1:"+result1+"  time:"+(end-start));        start = System.currentTimeMillis();        double result2 = findMedianSortedArrays2(nums1, nums2);        end = System.currentTimeMillis();        System.out.println("result 2:"+result2+"  time:"+(end-start));        start = System.currentTimeMillis();        double result3 = findMedianSortedArrays3(nums1, nums2);        end = System.currentTimeMillis();        System.out.println("result 3:"+result3+"  time:"+(end-start));    }

main中通过创建的俩5百万个数的数组对三种方式进行测试，得到计算时间如下（单位ms）

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