Leetcode -- Day 20 & Day 21

时间：2015-07-30 11:04:48 阅读：186 评论：0 收藏：0 [点我收藏+]

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Binary search in sorted Array

Question 1

Median of Two Sorted Arrays

There are two sorted arrays nums1 and nums2 of size m and n respectively. Find the median of the two sorted arrays. The overall run time complexity should be O(log (m+n)).

Refer to this link below:

http://www.cnblogs.com/springfor/p/3861890.html

 1 public class Solution {
 2     public double findMedianSortedArrays(int[] nums1, int[] nums2) {
 3         if (nums1 == null || nums2 == null)
 4             return 0;
 5         int m = nums1.length;
 6         int n = nums2.length;
 7         int k = (m+n)/2;
 8         
 9         if ((m+n) % 2 == 0){
10             double first = findMedianSortedArrays(nums1,0,m-1,nums2,0,n-1,k);
11             double second = findMedianSortedArrays(nums1,0,m-1,nums2,0,n-1,k+1);
12             return (first+second)/2;
13         }
14         else
15             return findMedianSortedArrays(nums1,0,m-1,nums2,0,n-1,k+1);
16     }
17     
18     public double findMedianSortedArrays(int[] A, int sA, int eA, int[] B, int sB, int eB, int k){
19         int m = eA - sA + 1;
20         int n = eB - sB + 1;
21         if (m > n)
22             return findMedianSortedArrays(B, sB, eB, A, sA, eA, k);
23         if (m == 0)
24             return B[k-1];
25         if (k == 1)
26             return Math.min(A[sA], B[sB]);
27         int partA = Math.min(k/2, m);
28         int partB = k - partA;
29         if (A[sA + partA-1] < B[sB + partB - 1])
30             return findMedianSortedArrays(A, sA + partA, eA, B, sB, eB, k-partA);
31         else if (A[sA + partA-1] > B[sB + partB - 1])
32             return findMedianSortedArrays(A, sA, eA, B, sB+partB, eB, k-partB);
33         else
34             return A[sA + partA - 1];
35     }
36 }

Search a 2D Matrix

Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties:

Integers in each row are sorted from left to right.
The first integer of each row is greater than the last integer of the previous row.

For example,

Consider the following matrix:

[
  [1,   3,  5,  7],
  [10, 11, 16, 20],
  [23, 30, 34, 50]
]

Given target = 3, return true.

row and colum‘s product -1 is the total number of this values. And a trick here is mid/colum is the row and mid%colum is the column.

public boolean searchMatrix(int[][] matrix, int target) {
        if (matrix == null || matrix.length < 1)
            return false;
        
        int left = 0;
        int right = matrix.length * matrix[0].length - 1;
        
        while (left <= right){
            int mid = (left + right)/2;
            int value = matrix[mid/matrix[0].length][mid % matrix[0].length];
            if (value == target){
                return true;
            }
            else if (value < target){
                left = mid + 1;
            }
            else {
                right = mid - 1;
            }
        }
        return false;
    }

This is my initial solution, using two loops, more complex but more directly.

 1 public boolean searchMatrix(int[][] matrix, int target) {
 2         if (matrix == null)
 3             return false;
 4             
 5         if (matrix[0][0] > target)
 6             return false;
 7         int row_mid;
 8         int col_mid;
 9         int row_start = 0;
10         int col_start = 0;
11         int row_end = matrix.length-1;
12         int col_end = matrix[0].length-1;
13         
14         while (row_start <= row_end){
15             row_mid = (row_start + row_end)/2;
16             // matrix[row_mid][] is the one we are interested in
17             if (matrix[row_mid][0] == target){
18                 return true;
19             }
20             else if (matrix[row_mid][0] > target){
21                 row_end = row_mid - 1; 
22             }
23             else {
24                 row_start = row_mid + 1;
25             }
26         }
27         
28         row_mid = row_end;
29         if (row_mid < 0)
30             return false;
31             
32         while (col_start <= col_end){
33             col_mid = (col_start + col_end)/2;
34             int value = matrix[row_mid][col_mid];
35             if (value == target){
36                 return true;
37             }
38             else if (value < target){
39                 col_start = col_mid + 1;
40             }
41             else{
42                 col_end = col_mid - 1;
43             }
44         }
45             
46         return false;
47     }

Leetcode -- Day 20 & Day 21

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原文地址：http://www.cnblogs.com/timoBlog/p/4688332.html

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