LeedCode第四题：寻找两个有序数组的中位数

Posted 2021-03-14 杯莫停、

这里我有两种解法，第一种执行时间是157ms，通过两个for循环排序（有点多余）效率比较慢：

class Solution {
    public double findMedianSortedArrays(int[] nums1, int[] nums2) {
        int len1 = nums1.length;
        int len2 = nums2.length;
        int i = 0;
        int j = 0;
        double res = 0;

        int[] nums3 = new int[len1 + len2];
        while (i < len1) {
            nums3[i] = nums1[i];
            i++;
        }

        while (j < len2) {
            nums3[len1] = nums2[j];
            len1++;
            j++;
        }

        int len3 = nums3.length;
        int temp = 0;
        for (int h = 0; h < len3; h++) {
            for (int k = 1; k < len3; k++) {
                if (nums3[k - 1] > nums3[k]) {
                    temp = nums3[k - 1];
                    nums3[k - 1] = nums3[k];
                    nums3[k] = temp;
                }
            }
        }
        if (len3 % 2 == 0) {
            int m = len3 / 2;
            res = (float)(nums3[m] + nums3[m - 1]) / 2;
        }
        if (len3 % 2 == 1) {
            int n = (len3 - 1) / 2;
            res = nums3[n];
        }
        return res;
    }
}

通过优化后的代码，执行时间是3ms，明显效率快很多：

class Solution {
    public double findMedianSortedArrays(int[] nums1, int[] nums2) {
        int len1 = nums1.length;
        int len2 = nums2.length;
        int i = 0;
        int j = 0;
        double res = 0;

        int[] nums3 = new int[len1 + len2];

        int len3 = nums3.length;
        int temp = 0;
        while(i < len1 && j < len2) {
            if (nums1[i] < nums2[j]) {
                nums3[temp] = nums1[i];
                temp++;
                i++;
            }
            else {
                nums3[temp] = nums2[j];
                temp++;
                j++;
            }
        }
        while (i < len1) {
            nums3[temp] = nums1[i];
            temp++;
            i++;
        }
        while (j < len2) {
            nums3[temp] = nums2[j];
            temp++;
            j++;
        }

        if (len3 % 2 == 0) {
            int m = len3 / 2;
            res = (float)(nums3[m] + nums3[m - 1]) / 2;
        }
        if (len3 % 2 == 1) {
            int n = (len3 - 1) / 2;
            res = nums3[n];
        }
        return res;
    }
}