再见小小ronnie的博客

/** * Two binary search. * When isLeft is true search the left index of the target, * else search the right index of the index. */ public class Solution { public static int helper(int[] num

public class Solution { public int mySqrt(int x) { long low = 0, high = x; while (low + 1 < high) { long mid = low + (high-low)/2; if (mid * mid > x)...

TreeMap is a Red-Black tree based NavigableMap implementation. The map is sorted according to the natural ordering of its keys. The time complexity for containsKey, get, put and remove operations ar

Math. Solve the quadratic function k/2(k+1) k public class Solution { public int arrangeCoins(int n) { // math return (int) (-1+Math.sqrt(1+8*(double)n))/2; } }Binary searc

Note that the param for guess() is the number you need to guess. e.g.  guess(5) = 1 when the number you need to guess is 6.  /** * We can easily figure out a O(n) solution which calls guess() from

public class Solution extends VersionControl { public int firstBadVersion(int n) { int low = 1, high = n, mid = 0; while (low + 1 < high) { mid = low + (high - low) / 2

/** * How quick sort works. * First choose a pivot, say the first element of the array, * set two pointer low and high point to the start and end of the array, * from high to low, find a element t

/** * Start from the top-right corner, check * if target > curr means we need to move to the next row b/c curr is the largest element in the row * if target < curr means we need to move to the prev

public class Solution { public boolean isPerfectSquare(int num) { long mid = 0; int low = 1, high = num; while (low <= high) { mid = (low+high)>>1;

public class Solution { public int searchInsert(int[] nums, int target) { int low = 0, high = nums.length-1, mid = 0; while (low <= high) { mid = (low+high) >> 1;

/** * Pigenhold principle and binary search. * All integers are between [1, n], * choose mid as (1+n)/2, then count the number of integers that are <= mid in the array * if the number is greater t