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Commit a718d3d

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sqrt x: done
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4 files changed

+98
-3
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4 files changed

+98
-3
lines changed

‎src/main/java/com/leetcode/arrays/binarysearch/PowXN.java

Lines changed: 1 addition & 1 deletion
Original file line numberDiff line numberDiff line change
@@ -63,7 +63,7 @@ public static double myPowNaive(double x, int n) {
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*/
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public static double myPow(double x, int n) {
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double res = 1;
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long absN = Math.abs((long) n);
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long absN = Math.abs((long) n);// used a long so that `absN / 2` doesn't overflow
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while (absN > 0) {
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if (absN % 2 == 1) res *= x; // store whatever we have calculated so far in the final result

‎src/main/java/com/leetcode/arrays/binarysearch/SearchInsertPosition.java

Lines changed: 2 additions & 1 deletion
Original file line numberDiff line numberDiff line change
@@ -26,7 +26,7 @@
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* Example 4:
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* Input: [1,3,5,6], 0
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* Output: 0
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*
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* <p>
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* Similar question: {@link SmallestLetterGreaterThanTarget}.
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*
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* @author rampatra
@@ -59,5 +59,6 @@ public static int searchInsert(int[] nums, int target) {
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public static void main(String[] args) {
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assertEquals(2, searchInsert(new int[]{1, 2}, 3));
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assertEquals(1, searchInsert(new int[]{1, 3, 5, 6}, 2));
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}
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}
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@@ -0,0 +1,62 @@
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package com.leetcode.arrays.binarysearch;
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import static org.junit.jupiter.api.Assertions.assertEquals;
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/**
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* Level: Easy
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* Link: https://leetcode.com/problems/sqrtx/
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* Description:
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* Implement int sqrt(int x).
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*
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* Compute and return the square root of x, where x is guaranteed to be a non-negative integer.
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*
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* Since the return type is an integer, the decimal digits are truncated and only the integer part
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* of the result is returned.
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*
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* Example 1:
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* Input: 4
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* Output: 2
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*
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* Example 2:
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* Input: 8
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* Output: 2
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* Explanation: The square root of 8 is 2.82842..., and since
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* the decimal part is truncated, 2 is returned.
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*
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* @author rampatra
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* @since 2019年08月19日
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*/
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public class SqrtX {
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/**
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* Runtime: <a href="https://leetcode.com/submissions/detail/253282868/">1 ms</a>.
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*
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* @param x
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* @return
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*/
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public static int mySqrt(int x) {
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if (x == 0 || x == 1) {
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return x;
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}
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long low = 1;
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long high = x / 2;
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while (low <= high) {
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long mid = low + (high - low) / 2;
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if (mid * mid == x) {
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return (int) mid;
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} else if (mid * mid < x) {
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low = mid + 1;
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} else {
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high = mid - 1;
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}
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}
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return (int) high;
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}
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public static void main(String[] args) {
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assertEquals(2, mySqrt(8));
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assertEquals(3, mySqrt(9));
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assertEquals(46339, mySqrt(2147395599));
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}
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}

‎src/main/java/com/leetcode/strings/AnagramsInString.java

Lines changed: 33 additions & 1 deletion
Original file line numberDiff line numberDiff line change
@@ -5,8 +5,40 @@
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import java.util.List;
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/**
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* Level: Easy
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* Level: Medium
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* Problem: https://leetcode.com/problems/find-all-anagrams-in-a-string/
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* Description:
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* Given a string s and a non-empty string p, find all the start indices of p's anagrams in s.
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*
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* Strings consists of lowercase English letters only and the length of both strings s and p will not be larger
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* than 20,100.
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*
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* The order of output does not matter.
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*
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* Example 1:
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*
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* Input:
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* s: "cbaebabacd" p: "abc"
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*
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* Output:
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* [0, 6]
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*
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* Explanation:
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* The substring with start index = 0 is "cba", which is an anagram of "abc".
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* The substring with start index = 6 is "bac", which is an anagram of "abc".
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*
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* Example 2:
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*
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* Input:
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* s: "abab" p: "ab"
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*
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* Output:
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* [0, 1, 2]
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*
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* Explanation:
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* The substring with start index = 0 is "ab", which is an anagram of "ab".
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* The substring with start index = 1 is "ba", which is an anagram of "ab".
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* The substring with start index = 2 is "ab", which is an anagram of "ab".
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*
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* @author rampatra
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* @since 2019年04月13日

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