遍历字符串s,然后遍历它,当遇到1时,就让j加1,记录1的个数,当遇到0时候,比较一下,是让1变0的次数j小,还是将0变化1的次数小,于是生成最优解dp
var minFlipsMonoIncr = function (s) {
var dp = 0;
var j = 0;
for (var i of s) {
i == "0" && (dp = Math.min(j, dp + 1));
i == "1" && j++;
}
return dp;
};
console.log(minFlipsMonoIncr("010110"));