题解 | #最长公共子串#

/**
 * longest common substring
 * @param str1 string字符串 the string
 * @param str2 string字符串 the string
 * @return string字符串
 */
function LCS( str1 ,  str2 ) {
    const dp = new Array(str1.length + 1);
    let max = 0;
    const map = new Map()
    for (let i = 0; i <= str1.length; i++) { // 初始化整个dp矩阵，每个值为0
      dp[i] = new Array(str2.length + 1).fill(0);
    } 
    
    for(let i = 1;i<=str1.length;i++){
        for(let j = 1;j<=str2.length;j++){
            if(str1[i-1] === str2[j-1]){
                dp[i][j] = dp[i - 1][j - 1] + 1
                max = Math.max(max,dp[i][j])
                if(!map.has(max)) map.set(max,i)  //避免重复
            }
        }
    }
    let startIndex = map.get(max) - max;
    let endIndex = map.get(max) ;
    return str1.substring(startIndex,endIndex)  //截取字符串
}
module.exports = {
    LCS : LCS
};