2020.7.10

   

3. 无重复字符的最长子串

第一次代码空间复杂度大：

public class Solution {
    public int lengthOfLongestSubstring(String s) {
        if(s.length() <= 0)
            return 0;
        StringBuffer strb = new StringBuffer(s.length());
        strb.append(s, 0, 1);
        int result = 1;
        int length = 1;
        for (int i = 1; i < s.length(); i++) {
            if(strb.indexOf(s.substring(i,i+1)) != -1){
                int index = strb.indexOf(s.substring(i,i+1));
                if(result < length)
                    result = length;
                strb = new StringBuffer(strb.substring(index + 1, strb.length()));
            }
            strb.append(s,i,i+1);
            length = strb.length();
        }
        return result > length ? result : length;
    }
}

第二次 滑动窗口+map

参考：https://leetcode-cn.com/problems/longest-substring-without-repeating-characters/solution/hua-jie-suan-fa-3-wu-zhong-fu-zi-fu-de-zui-chang-z/

public class Solution {
   public int lengthOfLongestSubstring(String s) {
        Map<Character, Integer> map = new HashMap<>();
        int n = s.length(), result = 0;
        int start = 0, end = 0;
        for (; end < n; end++) {
            char op = s.charAt(end);
            if(map.containsKey(op)){
                start = Math.max(start, map.get(op));
            }
            result = Math.max(result, end - start + 1);
            map.put(op, end + 1);
        }
        return result;
    }


}

4. 寻找两个正序数组的中位数

第一次，时间复杂度（m+n),空间复杂度(m+n)不符合

class Solution {
    public double findMedianSortedArrays(int[] nums1, int[] nums2) {
        int n = nums1.length + nums2.length;
        int[]num = new int[n+1];
        int i = 0, j = 0, k = 0;
        for (; i < n && j < nums1.length && k < nums2.length; i++) {
            if(nums1[j] < nums2[k]){
                num[i] = nums1[j++];
            }else{
                num[i] = nums2[k++];
            }
        }
        while(j < nums1.length){
            num[i++] = nums1[j++];
        }
        while(k < nums2.length){
            num[i++] = nums2[k++];
        }

        if(n % 2 == 0){
            return (double) (num[n/2 - 1] + num[n/2]) / 2;
        }else{
            return (double) num[n/2];
        }
    }
}

 

第二次：二分查找（太强了，我太菜了）

思路参考：https://leetcode-cn.com/problems/median-of-two-sorted-arrays/solution/xiang-xi-tong-su-de-si-lu-fen-xi-duo-jie-fa-by-w-2/

public double findMedianSortedArrays(int[] nums1, int[] nums2) {
    int n = nums1.length;
    int m = nums2.length;
    int left = (n + m + 1) / 2;
    int right = (n + m + 2) / 2;
    //将偶数和奇数的情况合并，如果是奇数，会求两次同样的 k 。
    return (getKth(nums1, 0, n - 1, nums2, 0, m - 1, left) + getKth(nums1, 0, n - 1, nums2, 0, m - 1, right)) * 0.5;  
}
    
    private int getKth(int[] nums1, int start1, int end1, int[] nums2, int start2, int end2, int k) {
        int len1 = end1 - start1 + 1;
        int len2 = end2 - start2 + 1;
        //让 len1 的长度小于 len2，这样就能保证如果有数组空了，一定是 len1（ 牛逼啊 ）
        if (len1 > len2) return getKth(nums2, start2, end2, nums1, start1, end1, k);
        if (len1 == 0) return nums2[start2 + k - 1];

        if (k == 1) return Math.min(nums1[start1], nums2[start2]);

        int i = start1 + Math.min(len1, k / 2) - 1;
        int j = start2 + Math.min(len2, k / 2) - 1;

        if (nums1[i] > nums2[j]) {
            return getKth(nums1, start1, end1, nums2, j + 1, end2, k - (j - start2 + 1));
        }
        else {
            return getKth(nums1, i + 1, end1, nums2, start2, end2, k - (i - start1 + 1));
        }
    }

 

 

 

 

 

 

标签：10,return,int,2020.7,length,num,nums1,nums2
来源： https://www.cnblogs.com/shish/p/13281511.html