LeetCode

12-IntegertoRoman.js

/**
 * @param {number} num
 * @return {string}
 */
var intToRoman = function(num) {
    var romanNums = ['I','V','X','L','C','D','M'];
    var ans = "";
    var nums = [];
    for(var v=(num%10); num>0; i++,v=(num%10)){
        nums.push(v);
        num -= v;
        num /= 10;
    }

    var len = nums.length;
    for(var i=0; i<len; i++){
        var currNum = nums[i];
        var romanPos = 2*i;
        var currRoman = "";
        
        if(currNum==4 || currNum==9){
            currRoman = romanNums[romanPos]+currRoman;
            currNum +=1;
        }
        if(currNum == 10){
            currRoman = currRoman+romanNums[romanPos+2];
            currNum -=10;
        }
        if(currNum>=5){
            currRoman = currRoman+romanNums[romanPos+1];
            currNum -=5;
        }
        
        if(currNum <= 3){
            for(; currNum>0; currNum--){
                currRoman = currRoman + romanNums[romanPos];
            }
        }
        ans = currRoman + ans;
    }
    return ans;
};

454-4sumII.js

/**
 * @param {number[]} A
 * @param {number[]} B
 * @param {number[]} C
 * @param {number[]} D
 * @return {number}
 */
//先兩倆合併在兩倆合併，降低時間複雜度
//別忘了要計算重複數
var fourSumCount = function(A, B, C, D) {
    var arr1 = {};
    var arr2 = {};
    var ans = 0;
    for(var i in A){
        for(var j in B){
            arr1[A[i]+B[j]] =  arr1[A[i]+B[j]]!==undefined?  arr1[A[i]+B[j]]+1:1;
        }
    }
    for(var i in C){
        for(var j in D){
            arr2[C[i]+D[j]] =  arr2[C[i]+D[j]]!==undefined?  arr2[C[i]+D[j]]+1:1;
        }
    }
    for(var i in arr1){
        ans = arr2[-1*parseInt(i)]!==undefined?ans+arr1[i]*arr2[-1*parseInt(i)]:ans;
    }
    return ans;
};