思路:类似桶排序,新设一个长度为n的数组,记录i出现的次数。找出出现次数超过1的元素
public int findRepeatNumber(int[] nums) {
int results[] = new int[nums.length];
for (int i=0;i<nums.length;i++){
results[nums[i]]++;
if (results[nums[i]]>1)
return nums[i];
}
return -1;
}
思路:出现排序数组,应该想到使用二分查找法。在mid=nums[mid]时,考虑好左右指针怎么移动。(结束条件应该是nums[left]=nums[mid]=num[right]才能标示出所有相同数字)
public int search(int[] nums, int target) {
if(nums.length==0)
return 0;
int left=0,right=nums.length-1;
if(target<nums[left] || target>nums[right])
return 0;
while(left<=right){
int mid = (left+right)/2;
if(target<nums[mid]){
right=mid-1;
}else if(target>nums[mid]){
left=mid+1;
}else{
if(nums[right]!=target){
right--;
}else if(nums[left]!=target){
left++;
}else{
break;
}
}
}
return right-left+1;
}
思路:出现排序数组想到二分查找。当mid!=nums[mid]时,证明缺失数字在左边;相等时在右边。注意指针变化
public int missingNumber(int[] nums) {
int left=0,right=nums.length-1;
int mid=(left+right)/2;
while(left<=right){
if(mid!=nums[mid]){
right = mid-1;
}else{
left = mid+1;
}
mid = (left+right)/2;
}
return left;
}