Iterate
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| class Solution { public int missingNumber(int[] nums) { for (int i = 0; i < nums.length; i++) { if (i != nums[i]) { return i; } }
return nums.length; } }
|
XOR
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| class Solution { public int missingNumber(int[] nums) { int res = 0;
for (int i = 0; i < nums.length; i++) { res ^= nums[i]; res ^= i; } res ^= nums.length;
return res; } }
|
Binary Search
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| class Solution { public int missingNumber(int[] nums) { int left = 0, right = nums.length - 1; while (left <= right) { int mid = (left + right) >>> 1; if (nums[mid] == mid) { left = mid + 1; } else { right = mid - 1; } }
return left; } }
|
References
剑指 Offer 53 - II. 0~n-1中缺失的数字
面试题53 - II. 0~n-1 中缺失的数字(二分法,清晰图解)