4-5

4. Median of Two Sorted Arrays
   思路
   构造两个等长的集合：
   search m 二分find
   易错点：全左或全右赋值时错误

public class MedianOfTwoSortedArrayOfDifferentLength {

    public double findMedianSortedArrays(int input1[], int input2[]) {
        //if input1 length is greater than switch them so that input1 is smaller than input2.
        if (input1.length > input2.length) {
            return findMedianSortedArrays(input2, input1);
        }
        int x = input1.length;
        int y = input2.length;

        int low = 0;
        int high = x;
        while (low <= high) {
            int partitionX = (low + high)/2;
            int partitionY = (x + y + 1)/2 - partitionX;

            //if partitionX is 0 it means nothing is there on left side. Use -INF for maxLeftX
            //if partitionX is length of input then there is nothing on right side. Use +INF for minRightX
            int maxLeftX = (partitionX == 0) ? Integer.MIN_VALUE : input1[partitionX - 1];
            int minRightX = (partitionX == x) ? Integer.MAX_VALUE : input1[partitionX];

            int maxLeftY = (partitionY == 0) ? Integer.MIN_VALUE : input2[partitionY - 1];
            int minRightY = (partitionY == y) ? Integer.MAX_VALUE : input2[partitionY];

            if (maxLeftX <= minRightY && maxLeftY <= minRightX) {
                //We have partitioned array at correct place
                // Now get max of left elements and min of right elements to get the median in case of even length combined array size
                // or get max of left for odd length combined array size.
                if ((x + y) % 2 == 0) {
                    return ((double)Math.max(maxLeftX, maxLeftY) + Math.min(minRightX, minRightY))/2;
                } else {
                    return (double)Math.max(maxLeftX, maxLeftY);
                }
            } else if (maxLeftX > minRightY) { //we are too far on right side for partitionX. Go on left side.
                high = partitionX - 1;
            } else { //we are too far on left side for partitionX. Go on right side.
                low = partitionX + 1;
            }
        }

        //Only we we can come here is if input arrays were not sorted. Throw in that scenario.
        throw new IllegalArgumentException();
    }

    public static void main(String[] args) {
        int[] x = {1, 3, 8, 9, 15};
        int[] y = {7, 11, 19, 21, 18, 25};

        MedianOfTwoSortedArrayOfDifferentLength mm = new MedianOfTwoSortedArrayOfDifferentLength();
        mm.findMedianSortedArrays(x, y);
    }
}

时间复杂度 Olog（min（m，n））

5.Longest Palindromic Substring
中心开花；
o（n2）

class Solution {
    public String longestPalindrome(String s) {
        int start = 0, end = 0;
        int len1, len2, len;
        for (int i = 0; i < s.length(); i++) {
            len1 = extendCorner(s, i, i);
            len2 = extendCorner(s, i, i+1);
            len = Math.max(len1, len2);
            if (len > end - start) {
                start = i - (len - 1)/2;
                end = i + len/2;
            }
        }
        return s.substring(start, end+1);
    }
    private int extendCorner(String s, int left, int right) {
        while (left >= 0 && right < s.length() && s.charAt(left) == s.charAt(right)) {
            left--;
            right++;
        }
        
        return right - left - 1;
    }
}