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题目一 解法 class Solution { public int findLengthOfLCIS(int[] nums) { if(nums.l
解法
class Solution {
public int findLengthOfLCIS(int[] nums) {
if(nums.length==1) return 1;
int fast = 1;
int tmp = 1;
int max = Integer.MIN_VALUE;
while(fast<nums.length){
if(nums[fast]>nums[fast-1]){
tmp++;
max = Math.max(max,tmp);
}else{
max = Math.max(max,tmp);
tmp = 1;
}
fast++;
}
return max;
}
}
解法
class Solution {
public boolean validPalindrome(String s) {
int left = 0;
int right = s.length()-1;
while(left<right){
if(s.charAt(left)==s.charAt(right)){
left++;
right--;
}else{
String tmp = s.substring(left, right + 1);
return validPalindrome(tmp,1,tmp.length()-1)||validPalindrome(tmp,0,tmp.length()-2);
}
}
return true;
}
public boolean validPalindrome(String s, int low, int high) {
for (int i = low, j = high; i < j; ++i, --j) {
char c1 = s.charAt(i), c2 = s.charAt(j);
if (c1 != c2) {
return false;
}
}
return true;
}
}
解法
class Solution {
public double findMaxAverage(int[] nums, int k) {
int w = nums.length-k;
int max = Integer.MIN_VALUE;
for(int i = 0;i<=w;i++){
int res = 0;
for(int j = 0;j<k;j++){
res = nums[i+j]+res;
}
max = Math.max(max,res);
}
double ans = (double)max/k;
return ans;
}
}
解法
class Solution {
public int findShortestSubArray(int[] nums) {
int[] n = new int[50001];
for(int i = 0;i<nums.length;i++){
n[nums[i]]+=1;
}
int max = Integer.MIN_VALUE;
ArrayList<Integer> list = new ArrayList<Integer>();
for(int i = 0;i<n.length;i++){
max = Math.max(n[i],max);
}
for(int i = 0;i<n.length;i++){
if(n[i]==max){
list.add(i);
}
}
int min = Integer.MAX_VALUE;
for(int i = 0;i<list.size();i++){
int res = list.get(i);
int left = 0;
int right = nums.length-1;
while(nums[left]!=res){
left++;
}
while(nums[right]!=res){
right--;
}
min = Math.min(min,right-left+1);
}
return min;
}
}
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