Summary Ranges(228)
Given a sorted integer array without duplicates, return the summary of its ranges.
For example, given [0,1,2,4,5,7], return ["0->2","4->5","7"].
Credits:
Special thanks to @jianchao.li.fighter for adding this problem and creating all test cases.
public class Solution {
public List<String> summaryRanges(int[] nums) {
List<String> list=new ArrayList();
for(int i=0;i<nums.length;i++){
int a=nums[i];
while(i+1<nums.length&&(nums[i+1]-nums[i])==1){
i++;
}
if(a!=nums[i]){
list.add(a+"->"+nums[i]);
}else{
list.add(a+"");
}
}
return list;
}
}
238.Product of Array Except Self
Given an array of n integers where n > 1, nums, return an array output such that output[i] is equal to the product of all the elements of nums except nums[i].
Solve it without division and in O(n).
For example, given [1,2,3,4], return [24,12,8,6].
Follow up:
Could you solve it with constant space complexity? (Note: The output array does not count as extra space for the purpose of space complexity analysis.)
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public class Solution {
public int[] productExceptSelf(int[] nums) {
int n = nums.length;
int[] res = new int[n];
res[0] = 1;
for (int i = 1; i < n; i++) {
res[i] = res[i - 1] * nums[i - 1];
}
int right = 1;
for (int i = n - 1; i >= 0; i--) {
res[i] *= right;
right *= nums[i];
}
return res;
}
}
思路：
最開始是想全部乘史煎，然后除以自身缔杉。但是有0的話行不通。
上述解法是對于第i哥元素屁倔。先求 1 * arr[0] * … * arr[i-1];
然后再乘以arr[n-1] * ..arr[i+1];
268.Missing Number
Given an array containing n distinct numbers taken from 0, 1, 2, ..., n, find the one that is missing from the array.
For example,
Given nums = [0, 1, 3] return 2.
Note:
Your algorithm should run in linear runtime complexity. Could you implement it using only constant extra space complexity?
Credits:
Special thanks to @jianchao.li.fighter for adding this problem and creating all test cases.
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思路 ：
求和相見。差值就是缺少的數(shù)字庆聘。另外一種是異或。a^a=0.
public class Solution {
public int missingNumber(int[] nums) {
int sum=0;
int n=nums.length;
for(int i=0;i<n;i++){
sum+=nums[i];
}
int pre=n*(n+1)/2;
return pre-sum==0?0:pre-sum;
}
}
解法2
The basic idea is to use XOR operation. We all know that abb =a, which means two xor operations with the same number will eliminate the number and reveal the original number.
In this solution, I apply XOR operation to both the index and value of the array. In a complete array with no missing numbers, the index and value should be perfectly corresponding( nums[index] = index), so in a missing array, what left finally is the missing number.
public int missingNumber(int[] nums) {
int xor = 0, i = 0;
for (i = 0; i < nums.length; i++) {
xor = xor ^ i ^ nums[i];
}
return xor ^ i;
}
283.Move Zeroes
Given an array nums, write a function to move all 0's to the end of it while maintaining the relative order of the non-zero elements.
For example, given nums = [0, 1, 0, 3, 12], after calling your function, nums should be [1, 3, 12, 0, 0].
Note:
- You must do this in-place without making a copy of the array.
- Minimize the total number of operations.
Credits:
Special thanks to @jianchao.li.fighter for adding this problem and creating all test cases.
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思路：盡可能少的移動氛谜，那就是一個元素最好只移動一次掏觉。將第一個非o的放在第一個位置，第二個非零的放在第二個位置值漫，然后剩下的全部設(shè)為0.
public class Solution {
public void moveZeroes(int[] nums) {
if(nums.length==0 ||nums==null) return ;
int pos=0;
for(int num:nums){
if(num!=0){
nums[pos]=num;
pos++;
}
}
for(;pos<nums.length;pos++){
nums[pos]=0;
}
return;
}
}
287. Find the Duplicate Number
Given an array nums containing n + 1 integers where each integer is between 1 and n (inclusive), prove that at least one duplicate number must exist. Assume that there is only one duplicate number, find the duplicate one.
Note:
- You must not modify the array (assume the array is read only).
- You must use only constant, O(1) extra space.
- Your runtime complexity should be less than
O(n2). - There is only one duplicate number in the array, but it could be repeated more than once.
Credits:
Special thanks to @jianchao.li.fighter for adding this problem and creating all test cases.
Subscribe to see which companies asked this question.
思路：類似于一個連表有環(huán)，求環(huán)的交叉點织盼。
The main idea is the same with problem Linked List Cycle II,https://leetcode.com/problems/linked-list-cycle-ii/. Use two pointers the fast and the slow. The fast one goes forward two steps each time, while the slow one goes only step each time. They must meet the same item when slow==fast. In fact, they meet in a circle, the duplicate number must be the entry point of the circle when visiting the array from nums[0]. Next we just need to find the entry point. We use a point(we can use the fast one before) to visit form begining with one step each time, do the same job to slow. When fast==slow, they meet at the entry point of the circle. The easy understood code is as follows.
public class Solution {
public int findDuplicate(int[] nums) {
if(nums.length>1){
int slow=nums[0];
int fast=nums[nums[0]];
while(slow!=fast){
slow=nums[slow];
fast=nums[nums[fast]];
}
fast=0;
while(fast!=slow){
fast=nums[fast];
slow=nums[slow];
}
return slow;
}
return -1;
}
}
