LeetCode刷题日记之找出数组中重复数字

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对于一段程序，用不同的方法写消耗时间差距还是很大的，下面是本题的逐渐改进过程。

遍历

let nums = [0, 1, 2, 3, 4, 11, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15];
let findRepeatNumber = function (nums) {
    let startIndex = 0;
    let result;
    while (startIndex < nums.length) {
        let start = nums[startIndex];
        for (let i = startIndex + 1; i < nums.length; i++) {
            if (start === nums[i]) {
                result = nums[i];
            }
        }
        startIndex++;
    }
    return result;
};

减少时间复杂度，用set减少遍历次数

let findRepeatNumber = function (nums) {
    let result;
    let single = new Set();
    for (let i = 0; i < nums.length; i++) {
        if (single.has(nums[i])) {
            return nums[i];
        } else {
            single.add(nums[i]);
        }
    }
    return result;
};

先排序，再使用方法二

let findRepeatNumber = function (nums) {
    let result;
    nums = nums.sort((a, b) => {
        return a - b;
    });
    let single = new Set();
    for (let i = 0; i < nums.length; i++) {
        if (single.has(nums[i])) {
            return nums[i];
        } else {
            single.add(nums[i]);
        }
    }
    return result;
};
console.log(findRepeatNumber(nums));