递归

目标和

// 广度便利
const findTrage = (nums,target)=> {
  const dfs = (deep,nums,target) => {
    if(deep === nums.length) {
        if(target === 0) res++;
        return;
    }
    dfs(deep+1,nums, target + nums[deep]);
    dfs(deep+1,nums, target - nums[deep])

  }
  let res = 0;
  dfs(0,nums,target);
  return res;
}

console.log(findTrage([1,1,1,1,1],3)

代码题青蛙跳

function frogJump(n) {
    if (n <= 1) {
        return 1;
    }
    let dp = new Array(n + 1);
    dp[0] = 1; // 0级台阶有1种跳法（不跳）
    dp[1] = 1; // 1级台阶有1种跳法（跳1级）
    for (let i = 2; i <= n; i++) {
        // 到达第i级台阶的跳法等于到达第i-1级台阶的跳法加上到达第i-2级台阶的跳法
        dp[i] = dp[i - 1] + dp[i - 2];
    }
    return dp[n];
}

// 示例
console.log(frogJump(3)); // 输出: 3