代码挑战：找到单一数字-Javascript
【摘要】 Given a non-empty array of integers, every element appears twice except for one. Find that single one.Note:Your algorithm should have a linear runtime complexity. Could you implement it without usi...
Given a non-empty array of integers, every element appears twice except for one. Find that single one.
Note:
Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?
Example 1:
Input: [2,2,1]
Output: 1
Example 2:
Input: [4,1,2,1,2]
Output: 4方案1：
var singleNumber = function(nums) {
let result = nums[0];
for(let i = 1; i < nums.length; i ++) {
result ^= nums[i];
}
return result;
};
方案2：
var singleNumber = function(nums) {
let set = Set();
for(let i = 0; i < nums.length; i ++) {
if(set.has(nums[i])) {
set.delete(nums[i]);
}
else {
set.add(nums[i]);
}
}
for (let item of set) {
return item;
}
return 0;
};
测试案例：
describe("singleNumber()", function () {
it("test1", function () {
// 1. ARRANGE
let nums = [2, 2, 1];
// 2. ACT
var result = singleNumber(nums);
// 3. ASSERT
expect(result).to.be.equal(1);
});
it("test2", function () {
// 1. ARRANGE
let nums = [4, 1, 2, 1, 2];
// 2. ACT
var result = singleNumber(nums);
// 3. ASSERT
expect(result).to.be.equal(4);
});
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