前端算法：给定一个整数数组和一个目标值，找出数组中和为目标值的两个数、判断一个整数是否是回文数

淡淡的烟草味﹌ 2022-05-12 17:14 181阅读 0赞

<script>
    function twoSum(nums, target) {
      const hash = {};
      for (let i = 0; i < nums.length; i++) {
        if (hash[nums[i]] !== undefined) {
          return [hash[nums[i]], i];
        } else {
          let newKey = target - nums[i]
          hash[newKey] = i;
        }
      }
      return [-1,-1]
    }
    let nums = [2, 7, 11, 15], target = 17;
    console.log(twoSum(nums,target));
    </script>

<script>
        function IsPalindrome(x){
            // 特殊情况：
            // 如上所述，当 x < 0 时，x 不是回文数。
            // 同样地，如果数字的最后一位是 0，为了使该数字为回文，
            // 则其第一位数字也应该是 0
            // 只有 0 满足这一属性
            if(x < 0 || (x % 10 == 0 && x != 0)) {
                return false;
            }
            let rev = 0;                    // reversed number
            while (x >= 10) {
                let cur = x % 10;
                rev = rev*10 + cur;
                if (x === rev) return true; // check before changing
                x = ~~(x / 10); // better replace for Math.floor(), bit operator '~' inverts all the bits in your number and in the process converts the number to an int
                if (x === rev) return true; // check after changing
                if (x < rev) return false;  // stop in case reversed number becomes greater than original value
            }
            return false; // the longest case (diff in the middle) // [77778777, 10004001] 
        }
        console.log(IsPalindrome(69699696))
    </script>